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

1. Formula
   Observation

   $y = x^{3/2}$

y

Dependent variable representing the function value.

Real numbers

1. Formula
   Observation

   $y = x^{3/2}$

x

Independent variable.

Real numbers

1. Formula
   Observation

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

$f(x)$

The specific function being analyzed for arc length.

Real numbers

1. Formula
   Observation

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

f'(x)

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

Real numbers

1. Formula
   Observation

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

$\int_{a}^{b}$

Definite integral operator with limits a and b.

Real numbers

1. Formula
   Observation

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

a

Lower limit of integration.

Real numbers

1. Formula
   Observation

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

b

Upper limit of integration.

Real numbers

1. Formula
   Observation

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

dx

Differential indicating integration with respect to x.

Real numbers

1. Formula
   Observation

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

$f(x)$

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

$x \ge 0$

1. Formula
   Observation

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

f'(x)

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

$x \ge 0$

1. Formula
   Observation

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

(f'(x))^2

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

$x \ge 0$

1. Formula
   Observation

   The instructor writes $u = 1 + \frac{9}{4}x$.

2. Audio
   Observation

   He says, "So if I say u is equal to 1 plus 9 over 4 x".

u

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

$u \ge 1$ on the transformed interval

1. Formula
   Observation

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

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

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

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

1. Formula
   Observation

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

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

1. Formula
   Observation

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

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.

1. f is differentiable on the interval of integration

2. the curve is written as $y=f(x)$

1. Audio
   Observation

   The instructor says, "Straight up use substitution. So u substitution".

2. 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.

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.

1. the integrand contains a composite expression that can be simplified by substitution

2. for a definite integral, the bounds must also be changed

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

1. Audio
   Observation

   He says, "And then we just have to change the bounds of integration".

2. Formula
   Observation

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

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.

1. the substitution formula $u(x)$ is known

2. the original integral is definite

1. U-substitution for a definite integral

1. Audio
   Observation

   He says the antiderivative of $u^{1/2}$ is $u^{3/2}$ divided by $3/2$, i.e. multiplied by $2/3$.

2. Formula
   Observation

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

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

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

1. U-substitution for a definite integral

1. Audio
   Observation

   He says, "we know how to apply the fundamental or the second fundamental theorem of calculus here to evaluate this definite integral".

2. Formula
   Observation

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

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

1. an antiderivative has been found

2. the integral is definite

1. Antiderivative of a power function

1. 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.

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

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.

1. The function is $f(x)=x^{3/2}$.

2. The worked interval is $x\in[0,32/9]$.

3. The clip begins after the full setup has already been written on the board.

1. $f(x)$
2. f'(x)
3. (f'(x))^2

1. 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 $x=0\to u=1$, $x=\frac{32}{9}\to u=9$.

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.

1. Use when the radicand $1+\frac{9}{4}x$ is replaced by a single variable.

2. Both the differential and the limits must be changed consistently.

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

1. Formula
   Observation

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

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 $u^{1/2}$.

1. Applied to the substituted integrand $\sqrt{u}$.

2. Evaluated as a definite integral from $u=1$ to $u=9$.

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

1. Audio
   Observation

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

2. Formula
   Observation

   The board shows $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$.

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 $\frac{4}{9}$.

1. Both bracketed terms contain the same factor $\frac{2}{3}$.

2. This step is arithmetic simplification, not a new calculus theorem.

1. Antiderivative of $\sqrt{u}$

1. Formula
   Observation

   The board shows $9^{3/2}$ and $1^{3/2}$, then the next line becomes $\frac{8}{27}\cdot 26$.

2. 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."

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 $\frac{8}{27}\cdot 26$.

1. Uses exact integer powers rather than decimal approximation.

2. Applies after factoring out $\frac{2}{3}$.

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

1. Formula
   Observation

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

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.

1. $f(x)=x^{3/2}$

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

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

For the specific function and interval shown on the board.

1. Formula
   Observation

   The board rewrites the integral as $\frac{4}{9}\int_{u=1}^{u=9}\sqrt{u}$\,du.

2. 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.

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

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

2. $dx=\frac{4}{9}$du

3. $x=0$ corresponds to $u=1$

4. $x=\frac{32}{9}$ corresponds to $u=9$

For the definite integral shown in this worked example.

1. Formula
   Observation

   The board shows $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$.

2. Audio
   Observation

   He computes $9^{3/2}=27$ and notes the lower term is 1.

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

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

2. $1^{3/2}=1$

For the specific definite integral in this example.

1. Formula
   Observation

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

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

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

1. $f(x)=x^{3/2}$.

2. (f'(x))^$2=\frac{9}{4}x$.

3. The original x-interval is [0,$32/9$].

4. $u=1+\frac{9}{4}x$ and $dx=\frac{4}{9}$du.

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

1. Formula
   Observation

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

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

1. The integrand after substitution is $\sqrt{u}=u^{1/2}$.

2. An antiderivative is $\frac{2}{3}u^{3/2}$.

3. The limits are $u=1$ and $u=9$.

For the definite integral displayed on the board.

1. Formula
   Observation

   The board successively shows $\frac{8}{27}\cdot 26$ and then the boxed $\frac{208}{27}$.

2. Audio
   Observation

   The speaker concludes, "So it's 208 over 27. And we are done."

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

1. The preceding substitution and antiderivative steps are valid.

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

3. $8\cdot 26=208$.

For the specific example on the board.

1. Formula
   Observation

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

1. Expression

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

   Identify the given function.

   Justification

   Given in the problem statement.

   Shown in the video
2. 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
3. Expression

   $$(f'(x))^2 = \frac{9}{4}x$$
   Explanation

   Square the derivative.

   Justification

   Algebraic simplification.

   Shown in the video
4. 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

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

1. 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.

2. Audio
   Observation

   The instructor narrates each algebraic rearrangement.

1. 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
2. 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
3. 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
4. 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

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

1. Formula
   Observation

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

2. Audio
   Observation

   He explains why the upper x-bound was chosen so that the u-bound becomes 9.

1. 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
2. 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

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

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

2. Audio
   Observation

   He explicitly invokes the second fundamental theorem of calculus and computes $9^{3/2}=27$.

1. 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
2. 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
3. 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
4. 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
5. 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
6. 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

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

1. Formula
   Observation

   The board already shows the substituted integral and antiderivative at the start of the clip.

2. Audio
   Observation

   The speaker narrates factoring out $\frac{2}{3}$, computing $27-1=26$, and then multiplying $8\cdot 26$ to get 208.

3. Diagram
   Observation

   A cursor points to the terms being simplified as each new line is written.

1. 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.

1. 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
2. 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
3. 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
4. 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
5. 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
6. 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
7. 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