Worked example: Definite integral by thinking about the function's graph | Khan Academy

1. Formula
   Observation

   The definite integral is written as \int_{-3}^{3} \sqrt{9-x^2} dx.

\int_{-3}^{3} \sqrt{9-x^2} dx

Definite integral of the square root of nine minus x squared from negative three to three.

x \in [-3, 3]

1. Diagram
   Observation

   A vertical axis is drawn and labeled 'y'.

y

Vertical coordinate axis for graphing the function.

Real numbers

1. Diagram
   Observation

   A horizontal axis is drawn and labeled 'x'.

x

Horizontal coordinate axis for graphing the function.

Real numbers

1. Formula
   Observation

   The function is written as y = f(x) = \sqrt{9-x^2}.

f(x)

The function being integrated, defined as the square root of nine minus x squared.

x \in [-3, 3]

1. Formula
   Observation

   The equation is written as y^2 + x^2 = 9.

y^2 + x^2 = 9

Equation of a circle centered at the origin with radius three.

(x, y) \in \mathbb{R}^2

1. Formula
   Observation

   Top-left board shows $\int_{-3}^{3} \sqrt{9-x^2}\,dx$.

$\int_{-3}^{3} \sqrt{9-x^2}\,dx$

The definite integral whose value is being found geometrically.

Integration variable $x$ runs from $-3$ to $3$.

1. Formula
   Observation

   Right side begins with $y=f(x)=\sqrt{9-x^2}$.

2. Audio
   Observation

   The principal square root selects the nonnegative branch of the displayed circle relation.

$f(x)=\sqrt{9-x^2}$

The upper semicircle function obtained from the positive square root.

Real-valued for $-3 \le x \le 3$.

1. Formula
   Observation

   Board shows $y^2+x^2=9$.

2. Audio
   Observation

   The presenter identifies the circle relation and its radius 3.

$y^2+x^2=9$

Equation of the full circle of radius 3 centered at the origin.

All real $(x,y)$ satisfying the equation.

1. Formula
   Observation

   Board writes "radius = 3".

2. Audio
   Observation

   The presenter identifies the circle relation and its radius 3.

$r=3$

Radius of the circle and of the semicircle region used for the area computation.

Positive real number.

1. Formula
   Observation

   Green writing shows $\pi 3^2 = 9\pi$, then $\frac{\pi 3^2}{2}=\frac{9\pi}{2}$.

2. Audio
   Observation

   The presenter uses the full-circle area and takes half to evaluate this integral.

$\frac{9\pi}{2}$

Area of the upper semicircle, equal to the value of the definite integral.

Nonnegative real area.

1. Audio
   Observation

   The presenter sets the square-root integral over the symmetric interval as the problem.

2. Formula
   Observation

   \int_{-3}^{3} \sqrt{9-x^2} dx

The video presents a definite integral problem that can be solved by interpreting the integrand as the graph of a function rather than using algebraic integration techniques.

1. The limits of integration are -3 and 3.

2. The integrand is \sqrt{9-x^2}.

1. Audio
   Observation

   The presenter suggests recognizing the graph instead of finding an antiderivative.

For this continuous nonnegative function on the ordered interval [-3,3], the definite integral equals the ordinary area below its graph. Recognize the upper semicircle and use its area. For integrable functions that change sign, use signed contributions rather than adding every area positively. The general conditions are editorial.

1. The displayed function is continuous and nonnegative on [-3,3].

2. For the general interpretation, the function must be integrable and the interval ordered.

1. Definite Integral Problem

1. Audio
   Observation

   The principal square root selects the nonnegative branch of the displayed circle relation.

2. Formula
   Observation

   Board contrasts $y=f(x)=\sqrt{9-x^2}$ with the squared relation leading to $y^2+x^2=9$.

The video distinguishes the full circle relation from the function given by the positive square root. Taking $\sqrt{9-x^2}$ means choosing the nonnegative $y$-values, so the graph is only the top half of the circle rather than both halves.

1. Working over the reals.

2. Using the principal (nonnegative) square root.

1. Formula
   Observation

   Right side shows $y^2=9-x^2$, then $y^2+x^2=9$, labeled "circle" and "radius = 3".

2. Audio
   Observation

   The presenter identifies the circle relation and its radius 3.

After squaring, the relation becomes $y^2+x^2=9$, which is the standard equation of a circle centered at the origin with radius 3. This full circle contains both the upper and lower branches, unlike the original square-root function.

1. Center at $(0,0)$.

2. Radius $3$.

1. Principal square root selects the upper branch

1. Audio
   Observation

   The presenter restricts the real square root to inputs where its radicand is nonnegative.

2. Formula
   Observation

   The radicand is $9-x^2$ in the displayed function.

For the square root to remain real-valued, the expression inside must be nonnegative. The video states that when $|x|>3$, the quantity $9-x^2$ becomes negative, so the principal square root is not defined there in the real setting.

1. Real-valued interpretation of the square root.

2. Principal square root convention.

1. Principal square root selects the upper branch

1. Audio
   Observation

   The presenter links the shaded region above the horizontal axis with the integral.

2. Diagram
   Observation

   The region under the pink semicircle and above the x-axis is shaded green.

The clip uses the geometric meaning of the definite integral: because the graph lies on or above the x-axis on $[-3,3]$, the integral equals the area of the shaded region between the curve and the axis.

1. Function is nonnegative on the interval.

2. Interval matches the visible support of the semicircle.

3. The displayed function is continuous, hence integrable, on the closed ordered interval.

1. Graph of $y=\sqrt{9-x^2}$ is an upper semicircle

1. Audio
   Observation

   The presenter draws the upper semicircle in the coordinate plane.

2. Diagram
   Observation

   Axes are marked at $-3$, $3$ on the x-axis and $3$ on the y-axis; a pink upper semicircle is drawn.

The function is graphed as the top half of a circle of radius 3 centered at the origin. Its endpoints are $(-3,0)$ and $(3,0)$, and its highest point is $(0,3)$.

1. Principal square root.

2. Real plane coordinates.

1. Principal square root selects the upper branch
2. The squared equation describes the full circle

1. Audio
   Observation

   The presenter uses the full-circle area and takes half to evaluate this integral.

2. Formula
   Observation

   Computation proceeds via $\pi 3^2=9\pi$ and then division by 2.

Once the integrand is recognized as the upper semicircle, the integral can be computed from elementary geometry: find the area of the full circle and halve it. No integration technique is needed.

1. Recognize the graph as a semicircle.

2. Use the real geometric area formula for a circle.

1. Graph of $y=\sqrt{9-x^2}$ is an upper semicircle
2. Definite integral interpreted as signed area under the graph

1. Audio
   Observation

   The principal square root selects the nonnegative branch of the displayed circle relation.

2. Formula
   Observation

   Board labels the square-root expression as "top of circle".

For real $x$, $y=\sqrt{9-x^2}$ gives the upper semicircle of $x^2+y^2=9$.

1. Square root denotes the principal nonnegative root.

2. Working in the real plane.

For all real $x$ with $-3\le x\le 3$.

1. Audio
   Observation

   The presenter restricts the real square root to inputs where its radicand is nonnegative.

If $|x|>3$, then $9-x^2<0$, so $\sqrt{9-x^2}$ is not real-valued under the principal-root convention.

1. Real-valued square root.

2. Principal square root convention.

For all real $x$ with $|x|>3$.

1. Audio
   Observation

   The presenter uses the full-circle area and takes half to evaluate this integral.

2. Formula
   Observation

   Final written result is $\frac{9\pi}{2}$.

$\displaystyle \int_{-3}^{3} \sqrt{9-x^2}\,dx = \frac{9\pi}{2}$.

1. Interpret the integral as area under the graph on $[-3,3]$.

2. Recognize the graph as the upper semicircle of radius 3.

Exact equality for the displayed definite integral.

1. Formula
   Observation

   y = \sqrt{9-x^2}

2. Formula
   Observation

   y^2 = 9-x^2

3. Formula
   Observation

   y^2 + x^2 = 9

4. Audio
   Observation

   Squaring and rearranging the function equation identifies the associated circle.

1. Expression

   $$y = \sqrt{9-x^2}$$
   Explanation

   Start with the given function.

   Justification

   Definition of the integrand.

   Shown in the video
2. Expression

   $$y^2 = 9-x^2$$
   Explanation

   Squaring gives the displayed circle relation. Retain y≥0 from the principal square root and x∈[-3,3]; reversing the squared equation alone would also admit the lower half.

   Justification

   Algebraic manipulation.

   Supplementary explanation
3. Expression

   $$y^2 + x^2 = 9$$
   Explanation

   Add x^2 to both sides.

   Justification

   Algebraic manipulation.

   Shown in the video
4. Expression

   $$x^2+y^2=9,\quad (0,0),\quad r=3$$
   Explanation

   Recognize the standard form of a circle's equation.

   Justification

   Standard conic section knowledge.

   Supplementary explanation

The relation y²+x²=9 is a circle of radius 3. Together with y≥0, it gives the original upper semicircle; the full circle alone is not equivalent to the original function.

1. Formula
   Observation

   Board shows $y=f(x)=\sqrt{9-x^2}$, then $y^2=9-x^2$, then $y^2+x^2=9$.

2. Audio
   Observation

   The principal square root selects the nonnegative branch of the displayed circle relation.

1. Expression

   $$y=\sqrt{9-x^2}$$
   Explanation

   Start with the given function.

   Justification

   Displayed on the board.

   Shown in the video
2. Expression

   $$y^2=9-x^2$$
   Explanation

   Square both sides to remove the radical.

   Justification

   Algebraic manipulation shown on the board.

   Shown in the video
3. Expression

   $$y^2+x^2=9$$
   Explanation

   Rearrange into standard circle form.

   Justification

   Add $x^2$ to both sides.

   Shown in the video
4. Expression

   $$y=\sqrt{9-x^2}\ \Rightarrow\ y\ge 0$$
   Explanation

   Because the original expression uses the principal square root, only the upper half of the circle is retained.

   Justification

   Explicitly stated in the audio and annotated as "top of circle".

   Shown in the video

The graph of $y=\sqrt{9-x^2}$ is the upper semicircle of $x^2+y^2=9$.

1. Audio
   Observation

   The presenter uses the full-circle area and takes half to evaluate this integral.

2. Formula
   Observation

   Green writing shows $\pi 3^2=9\pi$, then $\frac{\pi 3^2}{2}=\frac{9\pi}{2}$.

1. Expression

   $$\int_{-3}^{3} \sqrt{9-x^2}\,dx = \text{area under } y=\sqrt{9-x^2} \text{ on } [-3,3]$$
   Explanation

   Rewrite the integral as the shaded geometric area.

   Justification

   Stated directly in the audio while the region is shaded.

   Shown in the video
2. Expression

   $$\text{full circle area} = \pi r^2 = \pi(3)^2 = 9\pi$$
   Explanation

   Compute the area of the entire circle of radius 3.

   Justification

   Standard circle-area formula, spoken and written.

   Shown in the video
3. Expression

   $$\text{semicircle area} = \frac{9\pi}{2}$$
   Explanation

   Take half of the full circle because the graph is only the top half.

   Justification

   The source halves the full-circle area because the shaded region is the upper half.

   Shown in the video
4. Expression

   $$\int_{-3}^{3} \sqrt{9-x^2}\,dx = \frac{9\pi}{2}$$
   Explanation

   Substitute the semicircle area back as the value of the integral.

   Justification

   Conclusion written on the board.

   Shown in the video

The definite integral equals $\frac{9\pi}{2}$.

1. Audio
   Observation

   The problem is approached by drawing and identifying the function graph.

2. Formula
   Observation

   \int_{-3}^{3} \sqrt{9-x^2} dx

1. The first 105-second interval stops before the area calculation; the full continuation completes the value.

Evaluate \int_{-3}^{3} \sqrt{9-x^2} dx.

1. The integrand is \sqrt{9-x^2}.

2. The limits of integration are -3 and 3.

Find the value of the definite integral by interpreting it as an area.

1. Expression

   $$y = \sqrt{9-x^2}$$
   Explanation

   Define the function to graph.

   Justification

   Identify the integrand as a function.

   Shown in the video
2. Expression

   $$y^2 + x^2 = 9$$
   Explanation

   Rearrange the equation to recognize the geometric shape.

   Justification

   Algebraic manipulation to find the conic section.

   Shown in the video
3. Expression

   $$x^2+y^2=9,\quad y\ge0,\quad r=3$$
   Explanation

   Since y = \sqrt{9-x^2} implies y \ge 0, the graph is the upper half of the circle.

   Justification

   Principal square root is non-negative.

   Supplementary explanation

At this stage the integral is identified as the upper-semicircle area with radius 3; the continuation performs the calculation.

Check if the derived geometric shape matches the function's domain and range.

1. Formula
   Observation

   Problem statement is $\int_{-3}^{3} \sqrt{9-x^2}\,dx$.

2. Diagram
   Observation

   A coordinate plane with the upper semicircle and shaded region is drawn.

3. Audio
   Observation

   The presenter uses the full-circle area and takes half to evaluate this integral.

Find the exact value of $\displaystyle \int_{-3}^{3} \sqrt{9-x^2}\,dx$.

1. Integrand: $\sqrt{9-x^2}$.

2. Limits of integration: $-3$ to $3$.

3. Underlying circle relation: x²+y²=9; retain y≥0 for the original square-root graph.

4. Radius: $3$.

Compute the definite integral without using an antiderivative.

1. Expression

   $$y=\sqrt{9-x^2}$$
   Explanation

   Identify the integrand as a function of $x$.

   Justification

   Given on the board.

   Shown in the video
2. Expression

   $$x^2+y^2=9$$
   Explanation

   Square to recognize the underlying circle.

   Justification

   Shown algebraically on the board.

   Shown in the video
3. Expression

   $$y\ge 0$$
   Explanation

   Because the original expression is the principal square root, keep only the upper semicircle.

   Justification

   Explicitly discussed in the audio.

   Shown in the video
4. Expression

   $$\int_{-3}^{3} \sqrt{9-x^2}\,dx = \text{area of upper semicircle}$$
   Explanation

   Translate the integral into a geometric area problem.

   Justification

   Stated while shading the region.

   Shown in the video
5. Expression

   $$\frac{1}{2}\pi(3)^2 = \frac{9\pi}{2}$$
   Explanation

   Use the circle-area formula and halve it.

   Justification

   Written and spoken in the final computation.

   Shown in the video

$\displaystyle \frac{9\pi}{2}$

The answer matches the area of a semicircle of radius 3, which is half of $9\pi$.

1. Animation
   Observation

   A vertical line is drawn, followed by a horizontal line intersecting it, forming a Cartesian coordinate system.

1. y-axis

2. x-axis

1. Axes are drawn on the black screen.

1. The origin remains at the intersection.

Setting up the coordinate plane to graph the function.

1. Animation
   Observation

   The equations y = f(x) = \sqrt{9-x^2}, y^2 = 9-x^2, and y^2 + x^2 = 9 are written sequentially.

1. Function equation

2. Squared equation

3. Circle equation

1. Equations appear one by one below the integral.

1. The integral remains visible at the top left.

Demonstrating the algebraic steps to transform the function into the standard equation of a circle.

1. Diagram
   Observation

   Left side already has axes; right side shows the algebraic setup and labels "circle" and "radius = 3".

2. Formula
   Observation

   Annotation "top of circle" is added beneath the integral expression.

1. Integral $\int_{-3}^{3} \sqrt{9-x^2}\,dx$

2. Function line $y=f(x)=\sqrt{9-x^2}$

3. Derived equations $y^2=9-x^2$ and $y^2+x^2=9$

4. Labels "circle", "radius = 3", "top of circle"

5. Coordinate axes

1. The annotation "top of circle" is added under the integral.

2. Attention shifts from algebraic recognition toward drawing the graph.

1. The integral expression remains fixed at top left.

2. The circle equation and radius label remain visible on the right.

The visual setup establishes that the integrand is not the whole circle but its upper branch.

1. Animation
   Observation

   Tick marks appear at $-3$, $3$ on the x-axis and $3$ on the y-axis, then a pink arc is drawn from $(-3,0)$ through $(0,3)$ to $(3,0)$.

2. Audio
   Observation

   The presenter draws the upper semicircle in the coordinate plane.

1. x-axis and y-axis

2. Tick marks at $-3$, $3$, and $3$

3. Pink semicircular arc

1. Axis labels are added first.

2. Then the upper semicircle is drawn across the interval $[-3,3]$.

1. Center remains the origin.

2. Radius remains 3 throughout the drawing.

The animation converts the algebraic description into the geometric object whose area will be integrated.

1. Animation
   Observation

   Vertical green strokes fill the region between the pink semicircle and the x-axis.

2. Audio
   Observation

   The presenter links the shaded region above the horizontal axis with the integral.

1. Pink upper semicircle

2. x-axis segment from $-3$ to $3$

3. Green shaded region

1. The previously empty interior under the arc becomes filled with green shading.

1. The boundary curve stays the same.

2. The interval endpoints stay at $-3$ and $3$.

The shaded region is the geometric meaning of the definite integral.

1. Formula
   Observation

   Green writing appears: $\pi 3^2 = 9\pi$, then $\frac{\pi 3^2}{2} = \frac{9\pi}{2}$, boxed.

2. Formula
   Observation

   The original integral is completed as $\int_{-3}^{3} \sqrt{9-x^2}\,dx = \frac{9\pi}{2}$.

1. Circle-area expression $\pi 3^2$

2. Semicircle-area expression $\frac{\pi 3^2}{2}$

3. Boxed result $\frac{9\pi}{2}$

4. Completed integral equation

1. The full-circle area is written first.

2. Then it is divided by 2 to obtain the semicircle area.

3. Finally the result is attached to the original integral.

1. Radius remains 3.

2. The region remains the upper half of the circle.

The visual conclusion ties the shaded area directly to the exact value of the integral.

1. Audio
   Observation

   The presenter distinguishes the single-valued square-root graph from the full circle relation.

Assuming that because the algebraic manipulation leads to x^2 + y^2 = 9, the graph of y = \sqrt{9-x^2} is the entire circle.

The function y = \sqrt{9-x^2} only produces non-negative y-values, so its graph is strictly the upper semicircle, not the full circle.

1. Audio
   Observation

   The principal square root selects the nonnegative branch of the displayed circle relation.

2. Formula
   Observation

   Board separately shows $y=\sqrt{9-x^2}$ and $y^2+x^2=9$.

One might think $\sqrt{9-x^2}$ already represents the entire circle $x^2+y^2=9$.

The squared equation describes the full circle, but the original principal square root gives only the upper semicircle with $y\ge 0$.

1. Audio
   Observation

   The presenter restricts the real square root to inputs where its radicand is nonnegative.

One might assume $\sqrt{9-x^2}$ can be evaluated for every real $x$.

Over the reals, the radicand must be nonnegative, so the function exists only for $-3\le x\le 3$.

1. Audio
   Observation

   The presenter uses the full-circle area and takes half to evaluate this integral.

One might think evaluating a definite integral always requires finding an antiderivative.

A recognizable graph can simplify integration geometrically. In this example the continuous nonnegative integrand makes the integral equal the ordinary shaded area; for a sign-changing integrable function, account for signed contributions.

1. Audio
   Observation

   The proposed solution uses a geometric area interpretation of the integral.

The method uses the area interpretation here because the displayed integrand is continuous and nonnegative across the entire integration interval.

1. Formula
   Observation

   Transformation from y = \sqrt{9-x^2} to y^2 + x^2 = 9.

Recognizing the circle relation, while retaining the nonnegative square-root branch, identifies the geometry used to evaluate this example.

1. Audio
   Observation

   The principal square root selects the nonnegative branch of the displayed circle relation.

2. Formula
   Observation

   Board annotation "top of circle" appears under the square-root expression.

Recognizing the principal square root is what turns the full circle equation into the upper semicircle graph.

1. Formula
   Observation

   The board derives $x^2+y^2=9$ and then draws the corresponding semicircle.

The graph of $y=\sqrt{9-x^2}$ is contained in the full circle $x^2+y^2=9$ as its upper half.

1. Audio
   Observation

   The presenter links the shaded region above the horizontal axis with the integral.

2. Diagram
   Observation

   The region under the drawn semicircle is shaded.

Once the graph is identified, the definite integral is applied as the area of that geometric region.

1. Audio
   Observation

   The presenter uses the full-circle area and takes half to evaluate this integral.

The area interpretation enables the geometric method of evaluating the integral from the circle formula.

1. Audio
   Observation

   The presenter restricts the real square root to inputs where its radicand is nonnegative.

2. Diagram
   Observation

   The drawn semicircle spans exactly from $-3$ to $3$ on the x-axis.

The real domain restriction explains why the graph occupies only the interval $[-3,3]$.

1. Audio
   Observation

   The lesson proposes evaluating the integral from its graph.

1. Definite Integral Problem
2. Geometric evaluation

1. Formula
   Observation

   Algebraic derivation of the circle equation.

1. Deriving the Circle Equation
2. Confusing the Function Graph with the Full Circle

1. Audio
   Observation

   The principal square root selects the nonnegative branch of the displayed circle relation.

1. Principal square root selects the upper branch
2. The squared equation describes the full circle
3. Graph of $y=\sqrt{9-x^2}$ is an upper semicircle

1. Audio
   Observation

   The presenter restricts the real square root to inputs where its radicand is nonnegative.

1. Domain of $\sqrt{9-x^2}$ over the reals
2. Graph of $y=\sqrt{9-x^2}$ is an upper semicircle

1. Audio
   Observation

   The presenter links the shaded region above the horizontal axis with the integral.

1. Definite integral interpreted as signed area under the graph
2. Graph of $y=\sqrt{9-x^2}$ is an upper semicircle

1. Audio
   Observation

   The presenter uses the full-circle area and takes half to evaluate this integral.

1. Evaluate this integral using circle area instead of antiderivatives
2. Definite integral interpreted as signed area under the graph

1. Formula
   Observation

   Final boxed result is $\frac{9\pi}{2}$.

1. Evaluate this integral using circle area instead of antiderivatives
2. Value of the definite integral