Calculus / English

Worked example: arc length | Applications of definite integrals | AP Calculus BC | Khan Academy

Khan Academy · YouTube · 7:03

YouTube · Khan AcademyOriginal video

IN THIS VIDEO

01Arc Length Formula
02Power Rule for Derivatives
03Setting up the Arc Length Integral

FROM THE LEARNING NOTES

\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx

Arc Length Formula

SELECTED MOMENT

Setting up the Integral

SUGGESTED PREPARATION

Derivatives
Power Rule
Definite Integrals
Derivatives of power functions
Definite integrals
u-substitution
Fundamental Theorem of Calculus
Derivative of a power function
u-substitution in definite integrals
Power rule for integration
Exact evaluation of fractional powers

READ & KEEP

The explanation, unpacked.

Reviewed learning material · Video analysis · English

Read the full overview

This video segment demonstrates how to set up a definite integral to find the arc length of the curve $y = x^(3/2)$ over the interval [0, $32/9$ ]. The presenter first states the general arc length formula, then calculates the derivative of the given function using the power rule. Finally, the derivative is squared and substituted into the formula to establish the specific integral needed for the calculation. This 180-second whiteboard segment works one calculus example in full: it starts from the arc-length formula for $y=x^{3/2}$ , specializes it to $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}$ \,dx, then uses u-substitution with $u=1+\frac{9}{4}x$ . The instructor derives du= $\frac{9}{4}dx$ and $dx=\frac{4}{9}$ du, changes the bounds from $x=0$ and $x=32/9$ to $u=1$ and $u=9$ , rewrites the integral as $\frac{4}{9}\int_1^9\sqrt{u}$ \,du, integrates to $\frac{4}{9}[\frac{2}{3}u^{3/2}]_1^9$ , and evaluates the result to $\frac{208}{27}$ . This 63-second whiteboard clip captures the end of an AP Calculus BC arc-length worked example for $f(x)=x^{3/2}$ . The board already shows the derivative, the substitution $u=1+9x/4$ , the converted limits 1 and 9, and the transformed integral ( $4/9$ ) $\int$ ₁⁹√u du. The narration then focuses on simplifying the evaluated antiderivative: factor out $2/3$ , compute $9^{3/2}-1^{3/2}=27-1=26$ , multiply by $8/27$ , and finish with the boxed exact value $208/27$ .

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Problem Introduction 1:20Arc Length Formula 1:50Setting up the Integral 3:00Arc-length setup and choice of substitution 3:40Changing the definite-integral bounds 4:14Rewriting the integral in terms of u 5:00Antiderivative and final evaluation 6:00Review of the substituted arc-length integral 6:18Evaluate

9^{3/2}-1^{3/2}

6:32Multiply to the final boxed answer

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The video begins by presenting a graph of the function $y = x^{3/2}$ on a coordinate plane. The goal is to find the arc length of this curve over a specific interval.

The interval chosen is from $x = 0$ to $x = 32/9$ . A yellow highlight is drawn on the graph to visually represent the segment of the curve whose length is being calculated.

To solve this, the general formula for arc length is introduced: $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ . This formula requires finding the derivative of the function.

For the given function $f(x) = x^{3/2}$ , the power rule is applied to find its derivative, resulting in $f'(x) = \frac{3}{2}x^{1/2}$ .

Next, the derivative is squared to fit into the arc length formula: $(f'(x))^2 = \left(\frac{3}{2}x^{1/2}\right)^2 = \frac{9}{4}x$ .

Finally, these components are substituted back into the integral along with the limits of integration, setting up the definite integral $\int_{0}^{32/9} \sqrt{1 + \frac{9}{4}x} dx$ ready for evaluation.

The clip begins with the general arc-length formula already on the board: $L=\int_a^b \sqrt{1+(f'(x))^2}\,dx$ . For the specific curve $y=x^{3/2}$ , the derivative work has already produced $(f'(x))^2=\frac{9}{4}x$ , so the integral to evaluate is $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$ .

The instructor chooses the substitution $u=1+\frac{9}{4}x$ because it is exactly the expression under the radical. Differentiating gives $\frac{du}{dx}=\frac{9}{4}$ , hence $du=\frac{9}{4}dx$ and therefore $dx=\frac{4}{9}du$ .

Because this is a definite integral, the bounds must also be rewritten in terms of $u$ . Substituting $x=0$ gives $u=1$ , and substituting $x=\frac{32}{9}$ gives $u=1+\frac{9}{4}\cdot\frac{32}{9}=1+8=9$ .

With the new variable and new limits in place, the original integral becomes $\frac{4}{9}\int_{1}^{9}\sqrt{u}\,du$ . The factor $\frac{4}{9}$ comes directly from replacing $dx$ by $\frac{4}{9}du$ .

Now the integrand is a simple power: $\sqrt{u}=u^{1/2}$ . Its antiderivative is $\frac{u^{3/2}}{3/2}=\frac{2}{3}u^{3/2}$ , so the definite integral is evaluated as $\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{1}^{9}$ .

Plugging in the bounds gives $\frac{4}{9}\left(\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right)$ . Since $9^{3/2}=27$ and $1^{3/2}=1$ , this simplifies to $\frac{4}{9}\cdot\frac{2}{3}(27-1)=\frac{8}{27}\cdot 26=\frac{208}{27}$ .

At the start of the clip, the whiteboard already contains the full calculus setup: $f(x)=x^{3/2}$ , $f'(x)=\frac{3}{2}x^{1/2}$ , $(f'(x))^2=\frac{9}{4}x$ , and the substitution $u=1+\frac{9}{4}x$ with $dx=\frac{4}{9}du$ . The original limits $x=0$ and $x=\frac{32}{9}$ have already been converted to $u=1$ and $u=9$ , leaving the definite integral $\frac{4}{9}\int_{1}^{9}\sqrt{u}\,du$ .

The next visible step applies the power rule to $\sqrt{u}=u^{1/2}$ , giving $\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{1}^{9}$ . Evaluated at the bounds, this becomes $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$ .

The speaker now chooses an arithmetic shortcut: factor out the common $\frac{2}{3}$ . That rewrites the expression as $\frac{4}{9}\cdot\frac{2}{3}\left[9^{3/2}-1^{3/2}\right]$ . Multiplying the outside constants gives $\frac{8}{27}$ , so the problem reduces to evaluating the bracket.

Because $9^{3/2}=27$ and $1^{3/2}=1$ , the bracket is $27-1=26$ . The board therefore shows $\frac{8}{27}\cdot 26$ .

The final step is pure multiplication. The speaker computes $8\cdot 26$ by splitting it into $160+48=208$ , so the exact value becomes $\frac{208}{27}$ . This fraction is written on the board and boxed to mark the end of the worked example.

Knowledge cards

Arc Length Formula

The formula used to calculate the length of a curve defined by a function $f(x)$ over an interval [a, b].

\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx

Power Rule for Derivatives

A rule used to find the derivative of a function of the form $x^n$ .

\frac{d}{dx}(x^n) = nx^{n-1}

Setting up the Arc Length Integral

Demonstrates how to substitute a specific function and its derivative into the arc length formula to create a solvable definite integral.

Arc length formula used in the example

The video starts from the standard formula for the length of a curve written as $y=f(x)$ on $[a,b]$ : $L=\int_a^b \sqrt{1+(f'(x))^2}\,dx$ . In this example, $f(x)=x^{3/2}$ , so $(f'(x))^2=\frac{9}{4}x$ and the integral becomes $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$ .

L=\int_a^b \sqrt{1+\left(f'(x)\right)^2}\,dx

Choosing the substitution variable

The instructor sets $u=1+\frac{9}{4}x$ because that is the expression inside the square root. This turns the integrand into $\sqrt{u}$ , which is much easier to integrate.

u=1+\frac{9}{4}x

Differential conversion for u-substitution

Differentiating $u=1+\frac{9}{4}x$ gives $\frac{du}{dx}=\frac{9}{4}$ , so $du=\frac{9}{4}dx$ . Solving for $dx$ yields $dx=\frac{4}{9}du$ , which introduces the constant factor $\frac{4}{9}$ in the transformed integral.

du=\frac{9}{4}dx,\quad dx=\frac{4}{9}du

Changing the bounds from x to u

For a definite integral, the limits must be converted along with the variable. Here $x=0$ gives $u=1$ , and $x=\frac{32}{9}$ gives $u=9$ . That is why the transformed integral runs from 1 to 9.

x=0\Rightarrow u=1,\qquad x=\frac{32}{9}\Rightarrow u=9

Transformed integral after substitution

After substituting both the integrand and the differential, the original arc-length integral becomes $\frac{4}{9}\int_{1}^{9}\sqrt{u}\,du$ . The factor $\frac{4}{9}$ is not optional; it comes from $dx=\frac{4}{9}du$ .

\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx=\frac{4}{9}\int_1^9\sqrt{u}\,du

Antiderivative of the square-root function

The integrand $\sqrt{u}$ is rewritten as $u^{1/2}$ . Using the power rule for integration, an antiderivative is $\frac{u^{3/2}}{3/2}=\frac{2}{3}u^{3/2}$ .

\int u^{1/2}\,du=\frac{2}{3}u^{3/2}

Final evaluation of the definite integral

Applying the Fundamental Theorem of Calculus gives $\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_1^9$ . Since $9^{3/2}=27$ and $1^{3/2}=1$ , the value simplifies to $\frac{208}{27}$ .

\frac{4}{9}\left(\frac{2}{3}\cdot 27-\frac{2}{3}\cdot 1\right)=\frac{208}{27}

Arc-length setup visible on the board

The clip begins with the calculus setup already written out for $f(x)=x^{3/2}$ . The derivative is $f'(x)=\frac{3}{2}x^{1/2}$ , so $(f'(x))^2=\frac{9}{4}x$ . The arc-length integrand is therefore based on $\sqrt{1+\frac{9}{4}x}$ , and the rest of the board shows how this is converted into a simpler integral by substitution.

f(x)=x^{3/2},\quad f'(x)=\frac{3}{2}x^{1/2},\quad (f'(x))^2=\frac{9}{4}x

Substitution and limit conversion

To simplify the radical, the example sets $u=1+\frac{9}{4}x$ . Then $\frac{du}{dx}=\frac{9}{4}$ , so $du=\frac{9}{4}dx$ and $dx=\frac{4}{9}du$ . The original bounds convert as $x=0\mapsto u=1$ and $x=\frac{32}{9}\mapsto u=9$ . This yields the transformed definite integral $\frac{4}{9}\int_{1}^{9}\sqrt{u}\,du$ .

u=1+\frac{9}{4}x,\quad dx=\frac{4}{9}du,\quad \frac{4}{9}\int_{1}^{9}\sqrt{u}\,du

Antiderivative of the substituted integrand

Since $\sqrt{u}=u^{1/2}$ , the power rule gives an antiderivative $\frac{2}{3}u^{3/2}$ . Applied to the definite integral, the board writes $\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{1}^{9}$ , then evaluates it at the endpoints.

\int u^{1/2}\,du=\frac{2}{3}u^{3/2},\quad \frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{1}^{9}

Why factor out $2/3$ ?

After endpoint substitution, both terms contain the same coefficient $\frac{2}{3}$ . The speaker explicitly factors it out to make the arithmetic easier, turning $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$ into $\frac{4}{9}\cdot\frac{2}{3}\left[9^{3/2}-1^{3/2}\right]$ .

\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]=\frac{4}{9}\cdot\frac{2}{3}\left[9^{3/2}-1^{3/2}\right]

Evaluating the bracket

The remaining powers are exact integers: $9^{3/2}=27$ and $1^{3/2}=1$ . Therefore the bracket equals $27-1=26$ . Combined with the outside constant $\frac{4}{9}\cdot\frac{2}{3}=\frac{8}{27}$ , the expression becomes $\frac{8}{27}\cdot 26$ .

9^{3/2}=27,\quad 1^{3/2}=1,\quad 27-1=26,\quad \frac{8}{27}\cdot 26

Final arithmetic and boxed answer

The last step multiplies the numerator: $8\cdot 26=208$ . The speaker decomposes this as $160+48$ , confirming the product. The final exact value shown on the board is $\frac{208}{27}$ , which is boxed as the completed result of the worked example.

8\cdot 26=208,\quad \frac{8}{27}\cdot 26=\frac{208}{27}

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 27

y

Clear evidence

Shown in the video

Evidence

Formula
Observation
$y = x^{3/2}$

Symbol

Meaning

Dependent variable representing the function value.

Domain

Real numbers

x

Clear evidence

Shown in the video

Evidence

Formula
Observation
$y = x^{3/2}$

Symbol

Meaning

Independent variable.

Domain

Real numbers

$f(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
$f(x) = x^{3/2}$

Symbol

$f(x)$

Meaning

The specific function being analyzed for arc length.

Domain

Real numbers

f'(x)

Clear evidence

Shown in the video

Evidence

Formula
Observation
f'(x) = $\frac{3}{2}x^{1/2}$

Symbol

f'(x)

Meaning

The derivative of the function $f(x)$ .

Domain

Real numbers

$\int_{a}^{b}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$

Symbol

$\int_{a}^{b}$

Meaning

Definite integral operator with limits a and b.

Domain

Real numbers

a

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$

Symbol

Meaning

Lower limit of integration.

Domain

Real numbers

b

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$

Symbol

Meaning

Upper limit of integration.

Domain

Real numbers

dx

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$

Symbol

Meaning

Differential indicating integration with respect to x.

Domain

Real numbers

$f(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The whiteboard shows $f(x) = x^{3/2}$ .

Symbol

$f(x)$

Meaning

The function whose graph is used in the arc-length integral.

Domain

$x \ge 0$

f'(x)

Clear evidence

Shown in the video

Evidence

Formula
Observation
The whiteboard shows f'(x) = $\frac{3}{2}x^{1/2}$ .

Symbol

f'(x)

Meaning

The derivative of $f(x)$ , used inside the arc-length formula.

Domain

$x \ge 0$

(f'(x))^2

Clear evidence

Shown in the video

Evidence

Formula
Observation
The whiteboard shows (f'(x))^ $2 = \frac{9}{4}x$ .

Symbol

(f'(x))^2

Meaning

The square of the derivative, substituted into the arc-length integrand.

Domain

$x \ge 0$

u

Clear evidence

Shown in the video

Evidence

Formula
Observation
The instructor writes $u = 1 + \frac{9}{4}x$ .
Audio
Observation
He says, "So if I say u is equal to 1 plus 9 over 4 x".

Symbol

Meaning

Substitution variable for the expression under the square root in the arc-length integral.

Domain

$u \ge 1$ on the transformed interval

Knowledge points · 13

Arc Length Formula

Clear evidence

Shown in the video

Evidence

Formula
Observation
Arc length = $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$

Formula

Explanation

The formula used to calculate the length of a curve defined by a function $f(x)$ over an interval [a, b].

Formula

\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx

Conditions

$f(x)$ must be continuous on [a, b]
f'(x) must be continuous on [a, b]

Power Rule for Derivatives

Clear evidence

Shown in the video

Evidence

Formula
Observation
f'(x) = $\frac{3}{2}x^{1/2}$

Method

Explanation

A rule used to find the derivative of a function of the form $x^n$ .

Formula

\frac{d}{dx}(x^n) = nx^{n-1}

Arc length formula for a graph $y=f(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board displays Arc length = $\int_a^b \sqrt{1+(f'(x))^2}$ \,dx.

Formula

Explanation

For a curve given by $y=f(x)$ on an interval [a,b], the arc length is computed by integrating the square root of 1 plus the squared derivative of f.

Formula

L=\int_a^b \sqrt{1+\left(f'(x)\right)^2}\,dx

Conditions

f is differentiable on the interval of integration
the curve is written as $y=f(x)$

U-substitution for a definite integral

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor says, "Straight up use substitution. So u substitution".
Formula
Observation
He writes $u = 1 + \frac{9}{4}x$ , then $\frac{du}{dx}=\frac{9}{4}$ , then du= $\frac{9}{4}dx$ , then $dx=\frac{4}{9}$ du.

Method

Explanation

To evaluate the arc-length integral, the expression under the radical is replaced by a new variable u, the differential dx is rewritten in terms of du, and the limits are converted from x-values to u-values.

Formula

u=1+\frac{9}{4}x,\quad du=\frac{9}{4}dx,\quad dx=\frac{4}{9}du

Conditions

the integrand contains a composite expression that can be simplified by substitution
for a definite integral, the bounds must also be changed

Prerequisites

Arc length formula for a graph $y=f(x)$

Changing definite-integral bounds under substitution

Clear evidence

Shown in the video

Evidence

Audio
Observation
He says, "And then we just have to change the bounds of integration".
Formula
Observation
The board shows $x=0 \to u=1$ and $x=\frac{32}{9} \to u=9$ .

Method

Explanation

When using u-substitution in a definite integral, each original x-bound is substituted into $u=1+\frac{9}{4}x$ to obtain the corresponding u-bound.

Formula

x=0\Rightarrow u=1,\qquad x=\frac{32}{9}\Rightarrow u=9

Conditions

the substitution formula $u(x)$ is known
the original integral is definite

Prerequisites

U-substitution for a definite integral

Antiderivative of a power function

Clear evidence

Shown in the video

Evidence

Audio
Observation
He says the antiderivative of $u^{1/2}$ is $u^{3/2}$ divided by $3/2$ , i.e. multiplied by $2/3$ .
Formula
Observation
The board shows $\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{u=1}^{u=9}$ .

Formula

Explanation

The integrand $\sqrt{u}=u^{1/2}$ is integrated using the power rule, producing $\frac{2}{3}u^{3/2}$ .

Formula

\int u^{1/2}\,du=\frac{u^{3/2}}{3/2}=\frac{2}{3}u^{3/2}

Conditions

$u \ge 0$ for the real-valued square root form

Prerequisites

U-substitution for a definite integral

Evaluating a definite integral with an antiderivative

Clear evidence

Shown in the video

Evidence

Audio
Observation
He says, "we know how to apply the fundamental or the second fundamental theorem of calculus here to evaluate this definite integral".
Formula
Observation
The board evaluates $\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{u=1}^{u=9}$ .

Method

Explanation

After finding an antiderivative, the definite integral is evaluated by subtracting the antiderivative at the lower bound from its value at the upper bound.

Formula

\int_{1}^{9}\sqrt{u}\,du=\left[\frac{2}{3}u^{3/2}\right]_{1}^{9}

Conditions

an antiderivative has been found
the integral is definite

Prerequisites

Antiderivative of a power function

Arc-length integrand for $f(x)=x^{3/2}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The visible board already contains $f(x)=x^{3/2}$ , f'(x)= $\frac{3}{2}x^{1/2}$ , (f'(x))^ $2=\frac{9}{4}x$ , and the transformed integral $\frac{4}{9}\int_{1}^{9}\sqrt{u}$ \,du.

Uncertainties

The general arc-length formula itself is not spoken or written during this clip; only its specialized consequences are visible.

Definition

Explanation

This clip works with the arc-length integrand for the curve $y=x^{3/2}$ . The board shows that the derivative is f'(x)= $\frac{3}{2}x^{1/2}$ , so its square is $\frac{9}{4}x$ . The remaining integral is therefore built from $\sqrt{1+\frac{9}{4}x}$ , which is then rewritten by substitution as an integral in u.

Formula

\sqrt{1+(f'(x))^2}=\sqrt{1+\frac{9}{4}x}

Conditions

The function is $f(x)=x^{3/2}$ .
The worked interval is $x\in[0,32/9]$ .
The clip begins after the full setup has already been written on the board.

Prerequisites

$f(x)$
f'(x)
(f'(x))^2

Substitution $u=1+\frac{9}{4}x$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $u=1+\frac{9}{4}x$ , $\frac{du}{dx}=\frac{9}{4}$ , du= $\frac{9}{4}dx$ , $dx=\frac{4}{9}$ du, and the endpoint conversions $x=0\to u=1$ , $x=\frac{32}{9}\to u=9$ .

Method

Explanation

The integral is simplified by introducing a new variable u equal to the expression under the square root. Differentiating gives du= $\frac{9}{4}dx$ , hence $dx=\frac{4}{9}$ du. The original x-limits are converted to u-limits, producing a definite integral entirely in u.

Formula

u=1+\frac{9}{4}x,\quad dx=\frac{4}{9}du,\quad x=0\mapsto u=1,\quad x=\frac{32}{9}\mapsto u=9

Conditions

Use when the radicand $1+\frac{9}{4}x$ is replaced by a single variable.
Both the differential and the limits must be changed consistently.

Prerequisites

Arc-length integrand for $f(x)=x^{3/2}$
u
$\frac{du}{dx}$
dx
$x=0$ ,\ $x=\frac{32}{9}$
$u=1$ ,\ $u=9$

Antiderivative of $\sqrt{u}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{u=1}^{u=9}$ .

Formula

Explanation

After substitution, the integrand becomes $\sqrt{u}=u^{1/2}$ . The displayed antiderivative is $\frac{2}{3}u^{3/2}$ , which is the result of applying the power rule for integration to $u^{1/2}$ .

Formula

\int u^{1/2}\,du=\frac{2}{3}u^{3/2}

Conditions

Applied to the substituted integrand $\sqrt{u}$ .
Evaluated as a definite integral from $u=1$ to $u=9$ .

Prerequisites

Substitution $u=1+\frac{9}{4}x$
$\sqrt{u}$
$\frac{2}{3}u^{3/2}$

Factoring the common coefficient $\frac{2}{3}$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says, "Actually let's just factor out the two thirds. That makes it easier."
Formula
Observation
The board shows $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$ .

Method

Explanation

Once the antiderivative is evaluated at both endpoints, the same coefficient $\frac{2}{3}$ appears in each term. The speaker explicitly factors it out before multiplying by the outside constant $\frac{4}{9}$ .

Formula

\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]=\frac{4}{9}\cdot\frac{2}{3}\left[9^{3/2}-1^{3/2}\right]

Conditions

Both bracketed terms contain the same factor $\frac{2}{3}$ .
This step is arithmetic simplification, not a new calculus theorem.

Prerequisites

Antiderivative of $\sqrt{u}$

Endpoint evaluation $9^{3/2}-1^{3/2}=26$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $9^{3/2}$ and $1^{3/2}$ , then the next line becomes $\frac{8}{27}\cdot 26$ .
Audio
Observation
The speaker says, "And then we're going to have 27 minus 1 inside ... So 27 minus 1 is just going to be 26."

Method

Explanation

The upper endpoint contributes $9^{3/2}=27$ and the lower endpoint contributes $1^{3/2}=1$ . Their difference is 26. Combined with $\frac{4}{9}\cdot\frac{2}{3}=\frac{8}{27}$ , the expression becomes $\frac{8}{27}\cdot 26$ .

Formula

9^{3/2}=27,\quad 1^{3/2}=1,\quad 27-1=26,\quad \frac{4}{9}\cdot\frac{2}{3}=\frac{8}{27}

Conditions

Uses exact integer powers rather than decimal approximation.
Applies after factoring out $\frac{2}{3}$ .

Prerequisites

Factoring the common coefficient $\frac{2}{3}$

Claims and conditions · 6

Specialized arc-length integral for this example

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}$ \,dx as the specialized arc-length integral.

Proposition

Statement

For $f(x)=x^{3/2}$ , the arc-length integral on [0, $32/9$ ] becomes $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}$ \,dx.

Hypotheses

$f(x)=x^{3/2}$
f'(x)= $\frac{3}{2}x^{1/2}$
(f'(x))^ $2=\frac{9}{4}x$

Quantifiers

For the specific function and interval shown on the board.

Equivalent u-integral after substitution

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board rewrites the integral as $\frac{4}{9}\int_{u=1}^{u=9}\sqrt{u}$ \,du.
Audio
Observation
He says the definite integral from $u=1$ to $u=9$ of the square root of u, with dx replaced by $\frac{4}{9}$ du.

Proposition

Statement

Using $u=1+\frac{9}{4}x$ , the original definite integral equals $\frac{4}{9}\int_{1}^{9}\sqrt{u}$ \,du.

Hypotheses

$u=1+\frac{9}{4}x$
$dx=\frac{4}{9}$ du
$x=0$ corresponds to $u=1$
$x=\frac{32}{9}$ corresponds to $u=9$

Quantifiers

For the definite integral shown in this worked example.

Numerical evaluation of the transformed integral

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$ .
Audio
Observation
He computes $9^{3/2}=27$ and notes the lower term is 1.

Proposition

Statement

The evaluated expression simplifies to $\frac{208}{27}$ .

Hypotheses

$9^{3/2}=27$
$1^{3/2}=1$

Quantifiers

For the specific definite integral in this example.

Equivalent substituted definite integral

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board displays $\frac{4}{9}\int_{u=1}^{u=9}\sqrt{u}$ \,du.

Uncertainties

The clip does not restate the original arc-length formula in words; it only shows the already transformed integral.

Proposition

Statement

For the worked example on the board, the arc-length integral has been rewritten as $\frac{4}{9}\int_{1}^{9}\sqrt{u}$ \,du using $u=1+\frac{9}{4}x$ .

Hypotheses

$f(x)=x^{3/2}$ .
(f'(x))^ $2=\frac{9}{4}x$ .
The original x-interval is [0, $32/9$ ].
$u=1+\frac{9}{4}x$ and $dx=\frac{4}{9}$ du.

Quantifiers

For the specific definite integral shown on the board over $x\in[0,32/9]$ .

Evaluation of the substituted integral

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{u=1}^{u=9}$ and then $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$ .

Proposition

Statement

The substituted integral evaluates to $\frac{4}{9}\left(\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right)$ .

Hypotheses

The integrand after substitution is $\sqrt{u}=u^{1/2}$ .
An antiderivative is $\frac{2}{3}u^{3/2}$ .
The limits are $u=1$ and $u=9$ .

Quantifiers

For the definite integral displayed on the board.

Final numerical value of the worked integral

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board successively shows $\frac{8}{27}\cdot 26$ and then the boxed $\frac{208}{27}$ .
Audio
Observation
The speaker concludes, "So it's 208 over 27. And we are done."

Proposition

Statement

The value of the displayed definite integral is $\frac{208}{27}$ .

Hypotheses

The preceding substitution and antiderivative steps are valid.
$9^{3/2}=27$ and $1^{3/2}=1$ .
$8\cdot 26=208$ .

Quantifiers

For the specific example on the board.

Derivations and proofs · 5

Setting up the Arc Length Integral

Clear evidence

Shown in the video

Evidence

Formula
Observation
$\int_{0}^{32/9} \sqrt{1 + \frac{9}{4}x} dx$

Proof

Steps

Expression
$f(x) = x^{3/2}$
Explanation
Identify the given function.
Justification
Given in the problem statement.
Shown in the video
Expression
$f'(x) = \frac{3}{2}x^{1/2}$
Explanation
Find the derivative of the function using the power rule.
Justification
Power rule for derivatives.
Shown in the video
Expression
$(f'(x))^2 = \frac{9}{4}x$
Explanation
Square the derivative.
Justification
Algebraic simplification.
Shown in the video
Expression
$\int_{0}^{32/9} \sqrt{1 + \frac{9}{4}x} dx$
Explanation
Substitute the squared derivative and the limits of integration into the arc length formula.
Justification
Arc length formula.
Shown in the video

Conclusion

The definite integral required to find the arc length is set up as $\int_{0}^{32/9} \sqrt{1 + \frac{9}{4}x} dx$ .

Deriving the substitution relations

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board successively writes $u=1+\frac{9}{4}x$ , $\frac{du}{dx}=\frac{9}{4}$ , du= $\frac{9}{4}dx$ , $dx=\frac{4}{9}$ du.
Audio
Observation
The instructor narrates each algebraic rearrangement.

Proof

Steps

Expression
$u=1+\frac{9}{4}x$
Explanation
Choose u to be the expression under the square root.
Justification
This simplifies $\sqrt{1+\frac{9}{4}x}$ to $\sqrt{u}$ .
Shown in the video
Expression
$\frac{du}{dx}=\frac{9}{4}$
Explanation
Differentiate u with respect to x.
Justification
The derivative of 1 is 0 and the derivative of $\frac{9}{4}x$ is $\frac{9}{4}$ .
Shown in the video
Expression
$du=\frac{9}{4}dx$
Explanation
Rewrite the derivative relation in differential form.
Justification
Multiply both sides of $\frac{du}{dx}=\frac{9}{4}$ by dx.
Shown in the video
Expression
$dx=\frac{4}{9}du$
Explanation
Solve for dx so it can replace dx in the integral.
Justification
Multiply both sides by $\frac{4}{9}$ .
Shown in the video

Conclusion

The original x-integral can be rewritten in terms of u with $dx=\frac{4}{9}$ du.

Converting the x-bounds to u-bounds

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $x=0 \to u=1$ and $x=\frac{32}{9} \to u=9$ .
Audio
Observation
He explains why the upper x-bound was chosen so that the u-bound becomes 9.

Proof

Steps

Expression
$x=0 \Rightarrow u=1+\frac{9}{4}(0)=1$
Explanation
Substitute the lower x-bound into the substitution formula.
Justification
Direct evaluation of $u(x)$ .
Shown in the video
Expression
$x=\frac{32}{9} \Rightarrow u=1+\frac{9}{4}\cdot\frac{32}{9}=1+8=9$
Explanation
Substitute the upper x-bound into the substitution formula.
Justification
Direct evaluation of $u(x)$ .
Shown in the video

Conclusion

The definite integral bounds change from [0, $32/9$ ] in x to [1,9] in u.

Evaluating the transformed definite integral

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $\frac{4}{9}\int_{u=1}^{u=9}\sqrt{u}$ \,du, then $\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{u=1}^{u=9}$ , then $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$ .
Audio
Observation
He explicitly invokes the second fundamental theorem of calculus and computes $9^{3/2}=27$ .

Proof

Steps

Expression
$\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx=\frac{4}{9}\int_1^9\sqrt{u}\,du$
Explanation
Replace the integrand and differential using the substitution, and replace the bounds using the transformed limits.
Justification
u-substitution for definite integrals.
Shown in the video
Expression
$\sqrt{u}=u^{1/2}$
Explanation
Rewrite the square root as a fractional power.
Justification
Definition of the square-root power.
Shown in the video
Expression
$\int u^{1/2}\,du=\frac{2}{3}u^{3/2}$
Explanation
Find an antiderivative of $u^{1/2}$ .
Justification
Power rule for integration.
Shown in the video
Expression
$\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_1^9$
Explanation
Apply the Fundamental Theorem of Calculus to the definite integral.
Justification
Evaluate the antiderivative at the upper and lower bounds and subtract.
Shown in the video
Expression
$\frac{4}{9}\left(\frac{2}{3}\cdot 27-\frac{2}{3}\cdot 1\right)$
Explanation
Compute the powers at the bounds.
Justification
$9^{3/2}=27$ and $1^{3/2}=1$ .
Shown in the video
Expression
$\frac{4}{9}\cdot \frac{2}{3}(27-1)=\frac{8}{27}\cdot 26=\frac{208}{27}$
Explanation
Simplify the arithmetic to the final exact value.
Justification
Algebraic simplification.
Derived from the video

Conclusion

The arc-length integral evaluates to $\frac{208}{27}$ .

Simplification of the already substituted arc-length integral

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board already shows the substituted integral and antiderivative at the start of the clip.
Audio
Observation
The speaker narrates factoring out $\frac{2}{3}$ , computing $27-1=26$ , and then multiplying $8\cdot 26$ to get 208.
Diagram
Observation
A cursor points to the terms being simplified as each new line is written.

Uncertainties

The earlier derivation from the general arc-length formula to $\frac{4}{9}\int_{1}^{9}\sqrt{u}$ \,du is not spoken in this clip; it is only visible as prewritten board content.

Proof

Steps

Expression
$\frac{4}{9}\int_{1}^{9}\sqrt{u}\,du$
Explanation
Start from the substituted definite integral already written on the board.
Justification
Visible board state at the beginning of the clip.
Shown in the video
Expression
$=\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{1}^{9}$
Explanation
Replace the integrand by an antiderivative and keep the limits in u.
Justification
Power rule for integration applied to $u^{1/2}$ .
Shown in the video
Expression
$=\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$
Explanation
Evaluate the antiderivative at the upper and lower limits.
Justification
Fundamental theorem of calculus for a definite integral.
Shown in the video
Expression
$=\frac{4}{9}\cdot\frac{2}{3}\left[9^{3/2}-1^{3/2}\right]$
Explanation
Factor out the common coefficient $\frac{2}{3}$ from both terms inside the brackets.
Justification
Explicitly stated by the speaker: "let's just factor out the two thirds."
Shown in the video
Expression
$=\frac{8}{27}\left[27-1\right]$
Explanation
Compute $\frac{4}{9}\cdot\frac{2}{3}=\frac{8}{27}$ , and evaluate $9^{3/2}=27$ and $1^{3/2}=1$ .
Justification
Arithmetic simplification and exact power evaluation.
Shown in the video
Expression
$=\frac{8}{27}\cdot 26$
Explanation
Subtract inside the bracket.
Justification
Speaker says, "27 minus 1 is just going to be 26."
Shown in the video
Expression
$=\frac{208}{27}$
Explanation
Multiply the numerator 8 by 26.
Justification
Speaker decomposes $8\cdot 26$ as $160+48=208$ .
Shown in the video

Conclusion

The worked definite integral shown on the board simplifies to the boxed value $\frac{208}{27}$ .

Worked examples · 3

Finding the Arc Length of $y = x^(3/2)$

Clear evidence

Shown in the video

Evidence

Audio
Observation
find the arc length of this curve from when x equals 0 to when x is equal to 32 over 9
Formula
Observation
$y = x^{3/2}$

Problem

Find the arc length of the curve $y = x^(3/2)$ from $x = 0$ to $x = 32/9$ .

Given

$y = x^{3/2}$
Interval: [0, $32/9$ ]

Goal

Calculate the arc length of the specified curve over the given interval.

Steps

Expression
$\int_{0}^{32/9} \sqrt{1 + \frac{9}{4}x} dx$
Explanation
Set up the definite integral using the arc length formula.
Justification
Arc length formula and derivative calculation.
Shown in the video

Answer

The setup for the integral is $\int_{0}^{32/9} \sqrt{1 + \frac{9}{4}x} dx$ . The final evaluation is not shown in the clip.

Verification

Not shown in the clip.

Worked example: arc length of $y=x^{3/2}$ from $x=0$ to $x=32/9$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The full worked example is visible on the board from the arc-length setup through the final numerical expression.
Audio
Observation
The instructor narrates the substitution, bound change, antiderivative, and evaluation.

Problem

Compute the arc length of the curve $y=x^{3/2}$ on the interval [0, $32/9$ ].

Given

$f(x)=x^{3/2}$
f'(x)= $\frac{3}{2}x^{1/2}$
(f'(x))^ $2=\frac{9}{4}x$
interval endpoints $x=0$ and $x=\frac{32}{9}$

Goal

Evaluate $L=\int_0^{32/9}\sqrt{1+\frac{9}{4}x}$ \,dx.

Steps

Expression
$L=\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$
Explanation
Set up the arc-length integral using the displayed formula.
Justification
Arc length formula for $y=f(x)$ .
Shown in the video
Expression
$u=1+\frac{9}{4}x,\quad dx=\frac{4}{9}du$
Explanation
Use u-substitution to simplify the radical.
Justification
Differentiate u and solve for dx.
Shown in the video
Expression
$x=0\Rightarrow u=1,\quad x=\frac{32}{9}\Rightarrow u=9$
Explanation
Convert the definite-integral bounds.
Justification
Substitute the original x-bounds into $u(x)$ .
Shown in the video
Expression
$L=\frac{4}{9}\int_1^9\sqrt{u}\,du$
Explanation
Rewrite the whole integral in terms of u.
Justification
Substitution rule for definite integrals.
Shown in the video
Expression
$L=\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_1^9$
Explanation
Integrate $\sqrt{u}=u^{1/2}$ .
Justification
Power rule for integration and the Fundamental Theorem of Calculus.
Shown in the video
Expression
$L=\frac{4}{9}\left(\frac{2}{3}\cdot 27-\frac{2}{3}\cdot 1\right)=\frac{208}{27}$
Explanation
Evaluate at the bounds and simplify.
Justification
Arithmetic simplification using $9^{3/2}=27$ .
Derived from the video

Answer

$\frac{208}{27}$

Verification

The final board expression matches the substitution steps and the stated values $9^{3/2}=27$ and $1^{3/2}=1$ .

Worked arc-length example for $f(x)=x^{3/2}$ on [0, $32/9$ ]

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board presents a complete worked example with $f(x)=x^{3/2}$ , derivative data, substitution data, transformed integral, antiderivative, and final boxed answer.
Audio
Observation
The narration only covers the final algebraic simplification from the substituted integral to $\frac{208}{27}$ .

Uncertainties

The original problem statement is not spoken in this clip; it is inferred from the visible setup already on the board.

Problem

Evaluate the definite integral that arises from the arc-length setup for $f(x)=x^{3/2}$ over $x\in[0,32/9]$ , using the substitution already written on the board.

Given

$f(x)=x^{3/2}$ .
f'(x)= $\frac{3}{2}x^{1/2}$ .
(f'(x))^ $2=\frac{9}{4}x$ .
$u=1+\frac{9}{4}x$ .
$dx=\frac{4}{9}$ du.
$x=0\to u=1$ and $x=\frac{32}{9}\to u=9$ .
The transformed integral is $\frac{4}{9}\int_{1}^{9}\sqrt{u}$ \,du.

Goal

Compute the exact value of the displayed definite integral.

Steps

Expression
$\frac{4}{9}\int_{1}^{9}\sqrt{u}\,du$
Explanation
Begin with the substituted integral already on the board.
Justification
Visible starting state of the clip.
Shown in the video
Expression
$\frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_{1}^{9}$
Explanation
Integrate $\sqrt{u}=u^{1/2}$ .
Justification
Power rule for integration.
Shown in the video
Expression
$\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$
Explanation
Apply the limits of integration.
Justification
Definite-integral evaluation.
Shown in the video
Expression
$\frac{4}{9}\cdot\frac{2}{3}\left[9^{3/2}-1^{3/2}\right]$
Explanation
Factor out the common $\frac{2}{3}$ .
Justification
Explicitly narrated by the speaker.
Shown in the video
Expression
$\frac{8}{27}\cdot 26$
Explanation
Use $9^{3/2}=27$ , $1^{3/2}=1$ , and multiply the outside constants.
Justification
Arithmetic simplification shown on the board and spoken aloud.
Shown in the video
Expression
$\frac{208}{27}$
Explanation
Multiply 8 by 26 to obtain the final numerator.
Justification
Speaker computes $8\cdot 26$ as $160+48=208$ .
Shown in the video

Answer

$\frac{208}{27}$

Verification

The final value is boxed on the board, and the spoken arithmetic matches the displayed simplification $\frac{8}{27}\cdot 26=\frac{208}{27}$ .

Visual events · 7

Highlighting the Arc Segment

Clear evidence

Shown in the video

Evidence

Animation
Observation
A yellow line highlights the segment of the red curve between $x=0$ and $x=32/9$ .

Objects

Red curve
Yellow highlight line
Axes

Changes

A portion of the curve is highlighted in yellow to indicate the specific arc length being calculated.

Invariants

The overall shape of the curve remains unchanged.

Interpretation

The visual aid helps the viewer understand which part of the function's graph corresponds to the arc length problem being solved.

Persistent whiteboard layout

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A coordinate graph of $y=x^{3/2}$ appears on the left, while the arc-length formula and subsequent algebra fill the center and right of the screen.

Objects

coordinate axes
graph of $y=x^{3/2}$
arc-length formula
substitution work
evaluated antiderivative

Changes

New lines of algebra are written beneath the initial setup as the solution progresses.
The view scrolls downward around 29 seconds and again around 119 seconds to reveal more working space.

Invariants

The original graph and initial formulas remain visible above the newer work.
The problem stays focused on the same arc-length integral throughout.

Interpretation

The visual organization separates the geometric setup from the symbolic computation, making the substitution process easy to track step by step.

Scrolling to continue the derivation

Clear evidence

Shown in the video

Evidence

Animation
Observation
The canvas shifts downward to make room for writing $dx=\frac{4}{9}$ du.
Audio
Observation
The instructor says, "let me scroll down a little bit".

Objects

whiteboard canvas
existing equations

Changes

The visible region moves downward.
Additional blank space appears below the current equations.

Invariants

The mathematical content already written does not change.
The same problem continues without interruption.

Interpretation

The scroll is a workspace adjustment, not a conceptual transition; it allows the next algebraic line to be written clearly.

Second scroll before applying the Fundamental Theorem

Clear evidence

Shown in the video

Evidence

Animation
Observation
The canvas scrolls again to reveal more room for the antiderivative evaluation.

Objects

transformed integral
blank writing area

Changes

The lower portion of the board comes into view.
Space opens for writing $\frac{4}{9}[\frac{2}{3}u^{3/2}]_{u=1}^{u=9}$ .

Invariants

The transformed integral remains the same.
The substitution bounds remain 1 and 9.

Interpretation

This visual shift supports the transition from rewriting the integral to evaluating it exactly.

Color-coded whiteboard organization

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The black digital whiteboard is divided into three color-coded regions: yellow setup at upper left, cyan substitution work in the middle, and magenta integral evaluation on the right.

Objects

Yellow function and derivative formulas.
Cyan substitution formulas and limit conversions.
Magenta transformed integral and evaluation lines.
White cursor.

Changes

The rightmost magenta work expands downward as new simplification lines are added.
The cursor moves among the terms being discussed.

Invariants

The left and middle setup remain visible throughout the clip.
The overall problem stays focused on one definite integral.

Interpretation

The layout separates the original calculus setup, the substitution mechanics, and the final numerical evaluation, making the logical flow of the worked example visually explicit.

Cursor highlighting the common factor

Clear evidence

Shown in the video

Evidence

Animation
Observation
The cursor hovers near the two $\frac{2}{3}$ terms inside the bracket while the speaker discusses factoring them out.

Objects

Bracketed expression $\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}$
White cursor

Changes

Attention shifts from the whole bracket to the repeated coefficient $\frac{2}{3}$ .

Invariants

The underlying expression remains unchanged during this pointing phase.

Interpretation

The visual emphasis supports the algebraic step of extracting a common factor before multiplying constants.

Sequential writing of the simplification and boxed answer

Clear evidence

Shown in the video

Evidence

Animation
Observation
New lines appear sequentially beneath the magenta work: first $\frac{8}{27}$ , then $\cdot 26$ , then = $\frac{208}{27}$ , and finally a box around the answer.

Objects

Line $\frac{8}{27}$
Line $\frac{8}{27}\cdot 26$
Line = $\frac{208}{27}$
Rectangular box around the final fraction

Changes

The expression is rewritten in fewer terms as constants are combined.
The final fraction is enclosed to mark completion.

Invariants

All new lines are algebraically equivalent transformations of the previous expression.

Interpretation

The animation makes the arithmetic compression visible: combine coefficients, evaluate powers, subtract, multiply, and box the result.

Misconceptions · 4

Do not forget to change bounds in a definite u-substitution

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor explicitly says, "And then we just have to change the bounds of integration".
Formula
Observation
He then converts $x=0$ and $x=32/9$ into $u=1$ and $u=9$ .

Misconception

One might substitute $u=1+\frac{9}{4}x$ and rewrite dx but leave the original x-limits in place.

Clarification

For a definite integral, the limits must be rewritten in terms of the new variable; here they become 1 and 9.

The differential replacement contributes a constant multiplier

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows the factor $\frac{4}{9}$ pulled outside the integral.
Audio
Observation
He says, "I'm just going to take the $4/9$ and stick it out here".

Misconception

One might replace dx by $\frac{4}{9}$ du and forget that this introduces an overall factor of $\frac{4}{9}$ .

Clarification

Because $dx=\frac{4}{9}$ du, the transformed integral is $\frac{4}{9}\int_1^9\sqrt{u}$ \,du, not $\int_1^9\sqrt{u}$ \,du.

Substitution requires changing limits as well as the integrand

Clear evidence

Supplementary explanation

Evidence

Formula
Observation
The board explicitly changes both the integrand and the limits when moving from x to u.

Misconception

One might think u-substitution only replaces the expression under the radical and leaves the original x-limits in place.

Clarification

In this worked example, $x=0$ and $x=\frac{32}{9}$ are converted to $u=1$ and $u=9$ before evaluating the definite integral.

The differential contributes an outside constant

Clear evidence

Supplementary explanation

Evidence

Formula
Observation
The board shows $dx=\frac{4}{9}$ du and carries the outside factor $\frac{4}{9}$ into the transformed integral.

Misconception

A common error is to substitute u into the radical but forget the Jacobian factor coming from dx.

Clarification

Here du= $\frac{9}{4}dx$ , so $dx=\frac{4}{9}$ du, which produces the prefactor $\frac{4}{9}$ in front of the u-integral.

Concept relations · 9

Power Rule for Derivatives → Setting up the Arc Length Integral

Clear evidence

Shown in the video

Evidence

Formula
Observation
f'(x) = $\frac{3}{2}x^{1/2}$

Application

Explanation

The power rule is applied to find the derivative of the function, which is a necessary step in setting up the arc length integral.

Arc length formula for a graph $y=f(x) \to U$ -substitution for a definite integral

Clear evidence

Shown in the video

Evidence

Formula
Observation
The arc-length integral is immediately followed by the substitution $u=1+\frac{9}{4}x$ .

Application

Explanation

The general arc-length formula produces a definite integral that is then evaluated using u-substitution.

U-substitution for a definite integral → Changing definite-integral bounds under substitution

Clear evidence

Shown in the video

Evidence

Audio
Observation
After defining u, the instructor says the bounds must be changed.

Contains

Explanation

Changing the bounds is an essential substep of performing u-substitution on a definite integral.

U-substitution for a definite integral → Antiderivative of a power function

Clear evidence

Shown in the video

Evidence

Formula
Observation
After rewriting in terms of u, the integrand becomes $\sqrt{u}$ , whose antiderivative is then computed.

Application

Explanation

The substitution reduces the problem to a simpler power-function integral that can be handled by the power rule.

Antiderivative of a power function → Evaluating a definite integral with an antiderivative

Clear evidence

Shown in the video

Evidence

Audio
Observation
He explicitly mentions using the second fundamental theorem of calculus to evaluate the definite integral.

Proof dependency

Explanation

Once an antiderivative is found, the Fundamental Theorem of Calculus is what justifies evaluating the definite integral by endpoint subtraction.

Arc-length integrand for $f(x)=x^{3/2}$ → Substitution $u=1+\frac{9}{4}x$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The yellow derivative data leads to the cyan substitution $u=1+\frac{9}{4}x$ and then to the magenta integral in u.

Application

Explanation

The arc-length integrand motivates the specific substitution that turns $\sqrt{1+\frac{9}{4}x}$ into $\sqrt{u}$ .

Substitution $u=1+\frac{9}{4}x$ → Antiderivative of $\sqrt{u}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
After rewriting the integral as $\frac{4}{9}\int_{1}^{9}\sqrt{u}$ \,du, the next displayed step is $\frac{4}{9}[\frac{2}{3}u^{3/2}]_{1}^{9}$ .

Application

Explanation

Once the integral is expressed in u, the power rule can be applied directly to $u^{1/2}$ .

Antiderivative of $\sqrt{u}$ → Factoring the common coefficient $\frac{2}{3}$

Clear evidence

Shown in the video

Evidence

Audio
Observation
Immediately after the antiderivative line, the speaker says, "let's just factor out the two thirds."

Application

Explanation

Factoring out $\frac{2}{3}$ is the next simplification step after evaluating the antiderivative at both bounds.

Factoring the common coefficient $\frac{2}{3}$ → Final multiplication to $\frac{208}{27}$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board proceeds from the factored bracket to $\frac{8}{27}\cdot 26$ and then to $\frac{208}{27}$ .

Application

Explanation

After the common factor is removed, the remaining work is exact arithmetic evaluation of powers and products.

Find an answer · 10

What is the formula for calculating the arc length of a curve?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Arc length = $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$

Knowledge points

Arc Length Formula

Why choose $u=1+\frac{9}{4}x$ in this arc-length integral?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The instructor says, "Straight up use substitution".

Knowledge points

Arc length formula for a graph $y=f(x)$
U-substitution for a definite integral

How do the limits change when using u-substitution in a definite integral?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $x=0 \to u=1$ and $x=32/9 \to u=9$ .

Knowledge points

Changing definite-integral bounds under substitution

What is the antiderivative of $\sqrt{u}$ used here?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $\frac{2}{3}u^{3/2}$ as the antiderivative.

Knowledge points

Antiderivative of a power function

What is the exact value of the arc length in this worked example?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The final evaluated expression leads to $\frac{208}{27}$ .

Knowledge points

Evaluating a definite integral with an antiderivative
Worked example: arc length of $y=x^{3/2}$ from $x=0$ to $x=32/9$

Why does the presenter factor out $2/3$ before finishing the arc-length integral?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker explicitly says, "let's just factor out the two thirds. That makes it easier."

Knowledge points

Factoring the common coefficient $\frac{2}{3}$

How are the original x-limits converted to u-limits in this u-substitution example?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $x=0\to u=1$ and $x=\frac{32}{9}\to u=9$ .

Knowledge points

Substitution $u=1+\frac{9}{4}x$

Where does the outside factor $4/9$ come from after substituting $u=1+9x/4$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows du= $\frac{9}{4}dx$ and $dx=\frac{4}{9}$ du, followed by $\frac{4}{9}\int_{1}^{9}\sqrt{u}$ \,du.

Knowledge points

Substitution $u=1+\frac{9}{4}x$

Why does $9^{3/2}-1^{3/2}$ become 26 in this worked example?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board uses $9^{3/2}$ and then replaces the bracket by 27-1.

Knowledge points

Endpoint evaluation $9^{3/2}-1^{3/2}=26$

What is the final exact value obtained for the displayed arc-length integral?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The final boxed value on the board is $\frac{208}{27}$ .
Audio
Observation
The speaker says, "So it's 208 over 27. And we are done."

Knowledge points

Final multiplication to $\frac{208}{27}$
Worked arc-length example for $f(x)=x^{3/2}$ on [0, $32/9$ ]

Coverage and review notes

Covered · Introduction to the problem, identifying the function and the interval for arc length calculation.

Covered · Stating the general arc length formula.

Covered · Applying the formula to the specific function, calculating the derivative, and setting up the definite integral.

Covered · The arc-length setup and the u-substitution relations are introduced and derived.

Covered · The original x-bounds are converted to u-bounds.

Covered · The integral is rewritten fully in terms of u with the constant factor extracted.

Covered · The antiderivative is found, the Fundamental Theorem is applied, and the exact value is computed.

Covered · The board already shows the setup and substituted integral; the audio focuses on factoring out $2/3$ .

Covered · The speaker evaluates $9^{3/2}-1^{3/2}=26$ and writes the intermediate product $8/27 * 26$ .

Covered · The speaker computes $8*26=208$ , writes $208/27$ , boxes it, and ends the example.

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Reviewed subject paths

Arc length

Questions this video answers

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Understand why

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Understand why

Meet the concept

Meet the concept

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Understand why

Worked example: arc length | Applications of definite integrals | AP Calculus BC | Khan Academy

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Arc Length Formula

Power Rule for Derivatives

Setting up the Arc Length Integral

Arc length formula used in the example

Choosing the substitution variable

Differential conversion for u-substitution

Changing the bounds from x to u

Transformed integral after substitution

Antiderivative of the square-root function

Final evaluation of the definite integral

Arc-length setup visible on the board

Substitution and limit conversion

Antiderivative of the substituted integrand

Why factor out 2/32/32/3?

Evaluating the bracket

Final arithmetic and boxed answer

Detailed learning notes

y

x

f(x)f(x)f(x)

f'(x)

∫ab\int_{a}^{b}∫ab​

a

b

dx

f(x)f(x)f(x)

f'(x)

(f'(x))^2

u

Arc Length Formula

Power Rule for Derivatives

Arc length formula for a graph y=f(x)y=f(x)y=f(x)

U-substitution for a definite integral

Changing definite-integral bounds under substitution

Antiderivative of a power function

Evaluating a definite integral with an antiderivative

Arc-length integrand for f(x)=x3/2f(x)=x^{3/2}f(x)=x3/2

Substitution u=1+94xu=1+\frac{9}{4}xu=1+49​x

Antiderivative of u\sqrt{u}u​

Factoring the common coefficient 23\frac{2}{3}32​

Endpoint evaluation 93/2−13/2=269^{3/2}-1^{3/2}=2693/2−13/2=26

Specialized arc-length integral for this example

Equivalent u-integral after substitution

Numerical evaluation of the transformed integral

Equivalent substituted definite integral

Evaluation of the substituted integral

Final numerical value of the worked integral

Setting up the Arc Length Integral

Deriving the substitution relations

Converting the x-bounds to u-bounds

Evaluating the transformed definite integral

Simplification of the already substituted arc-length integral

Finding the Arc Length of y=x(3/2)y = x^(3/2)y=x(3/2)

Worked example: arc length of y=x3/2y=x^{3/2}y=x3/2 from x=0x=0x=0 to x=32/9x=32/9x=32/9

Worked arc-length example for f(x)=x3/2f(x)=x^{3/2}f(x)=x3/2 on [0,32/932/932/9]

Highlighting the Arc Segment

Persistent whiteboard layout

Scrolling to continue the derivation

Second scroll before applying the Fundamental Theorem

Color-coded whiteboard organization

Cursor highlighting the common factor

Sequential writing of the simplification and boxed answer

Do not forget to change bounds in a definite u-substitution

The differential replacement contributes a constant multiplier

Substitution requires changing limits as well as the integrand

The differential contributes an outside constant

Power Rule for Derivatives → Setting up the Arc Length Integral

Arc length formula for a graph y=f(x)→Uy=f(x) \to Uy=f(x)→U-substitution for a definite integral

U-substitution for a definite integral → Changing definite-integral bounds under substitution

U-substitution for a definite integral → Antiderivative of a power function

Antiderivative of a power function → Evaluating a definite integral with an antiderivative

Arc-length integrand for f(x)=x3/2f(x)=x^{3/2}f(x)=x3/2 → Substitution u=1+94xu=1+\frac{9}{4}xu=1+49​x

Substitution u=1+94xu=1+\frac{9}{4}xu=1+49​x → Antiderivative of u\sqrt{u}u​

Antiderivative of u\sqrt{u}u​ → Factoring the common coefficient 23\frac{2}{3}32​

Factoring the common coefficient 23\frac{2}{3}32​ → Final multiplication to 20827\frac{208}{27}27208​

Why factor out $2/3$ ?

$f(x)$

$\int_{a}^{b}$

$f(x)$

Arc length formula for a graph $y=f(x)$

Arc-length integrand for $f(x)=x^{3/2}$

Substitution $u=1+\frac{9}{4}x$

Antiderivative of $\sqrt{u}$

Factoring the common coefficient $\frac{2}{3}$

Endpoint evaluation $9^{3/2}-1^{3/2}=26$

Finding the Arc Length of $y = x^(3/2)$

Worked example: arc length of $y=x^{3/2}$ from $x=0$ to $x=32/9$

Worked arc-length example for $f(x)=x^{3/2}$ on [0, $32/9$ ]

Arc length formula for a graph $y=f(x) \to U$ -substitution for a definite integral

Arc-length integrand for $f(x)=x^{3/2}$ → Substitution $u=1+\frac{9}{4}x$

Substitution $u=1+\frac{9}{4}x$ → Antiderivative of $\sqrt{u}$

Antiderivative of $\sqrt{u}$ → Factoring the common coefficient $\frac{2}{3}$

Factoring the common coefficient $\frac{2}{3}$ → Final multiplication to $\frac{208}{27}$

Why choose $u=1+\frac{9}{4}x$ in this arc-length integral?

What is the antiderivative of $\sqrt{u}$ used here?

Why does the presenter factor out $2/3$ before finishing the arc-length integral?

Where does the outside factor $4/9$ come from after substituting $u=1+9x/4$ ?

Why does $9^{3/2}-1^{3/2}$ become 26 in this worked example?