Finding definite integrals using area formulas | AP Calculus AB | Khan Academy

1. Formula
   Observation

   The left side lists four integrals of the form ∫ f(x) dx.

2. Diagram
   Observation

   The graph on the right is labeled with the function name f.

f

The given function whose graph is used to evaluate definite integrals.

The displayed graph extends from−6 to6; this initial section works with the first two intervals.

1. Formula
   Observation

   Each integral is written with dx.

2. Diagram
   Observation

   The horizontal axis is labeled x.

x

Independent variable / horizontal coordinate on the graph.

Real variable along the x-axis.

1. Diagram
   Observation

   The vertical axis is labeled y.

y

Dependent value of the function f at x.

Real values shown on the vertical axis.

1. Audio
   Observation

   The circle-area computation uses the constant π.

2. Formula
   Observation

   The handwritten expression uses π in (π·2²)/2 = 2π.

π

Circle constant used in the area formula for a circle.

Positive real constant.

1. Audio
   Observation

   The upper curved region is identified as a semicircle with its shown radius.

2. Formula
   Observation

   The written calculation substitutes r = 2 into πr².

r

Radius of the circle associated with the semicircular region.

Positive real number; here r = 2.

1. Formula
   Observation

   The integrals are written as $\int f(x)\,dx$, and the graph is labeled $f$.

f(x)

The function whose definite integrals are being evaluated from its graph.

Piecewise graph on the displayed coordinate plane; values are positive above the x-axis and negative below it.

1. Diagram
   Observation

   Horizontal axis labeled $x$ with tick marks from about $-6$ to $6$.

2. Formula
   Observation

   Each integral contains $dx$.

x

Independent variable / horizontal coordinate used as the integration variable.

Real interval shown on the graph, approximately $[-6,6]$.

1. Diagram
   Observation

   Vertical axis labeled $y$ with tick marks from about $-5$ to $5$.

y

Dependent variable / vertical coordinate representing the value of $f(x)$.

Displayed range approximately $[-5,5]$.

1. Formula
   Observation

   Four definite integrals are listed: $\int_{-6}^{-2} f(x)\,dx$, $\int_{-2}^{1} f(x)\,dx$, $\int_{1}^{4} f(x)\,dx$, $\int_{4}^{6} f(x)\,dx$.

\int_a^b f(x)\,dx

Definite integral of $f$ from $a$ to $b$, interpreted here as signed area between the graph and the x-axis.

$a,b$ are the displayed limits of integration.

1. Formula
   Observation

   Circle-area expression written as $\pi \cdot 1^2$ and then divided by 2.

2. Audio
   Observation

   The circle-area formula is applied with the final semicircle radius, and the full-circle area is halved.

\pi

Constant in the area formula for a circle.

Positive real constant.

1. Audio
   Observation

   The circle-area formula is applied with the final semicircle radius, and the full-circle area is halved.

2. Formula
   Observation

   Written expression uses $1^2$ after substituting radius 1.

r

Radius of the circle used in the area formula $\pi r^2$.

Here $r=1$.

1. Audio
   Observation

   The first positive region is linked to the definite integral by its geometric area.

2. Diagram
   Observation

   The region under the semicircle from x=-6 to x=-2 is shaded to match the first integral.

For the first example, the video interprets $\int_{-6}^{-2} f(x)\,dx$ as the geometric area between the curve and the x-axis when the graph lies above the axis.

1. The function is above the x-axis on the interval being considered.

2. For a Riemann-integrable function on an ordered finite interval a<b, the definite integral is net signed area: regions above the axis add, while regions below subtract. The displayed piecewise curve is continuous, which is sufficient for integrability. The curved pieces are the semicircles specified in this geometric exercise, rather than arbitrary curves inferred to be circular.

1. Audio
   Observation

   The upper curved region is identified as a semicircle with its shown radius.

2. Formula
   Observation

   The handwritten work shows (π·2²)/2 = 2π.

The video computes the area of a semicircle by taking half of the full circle area πr². With r=2, it obtains (π·2²)/2 = 2π.

1. The region is exactly a semicircle.

2. The radius is known.

1. Definite integral as geometric area above the x-axis

1. Audio
   Observation

   The explanation identifies the lower region and assigns a negative sign to its integral contribution.

2. Diagram
   Observation

   The second shaded region lies below the x-axis between x=-2 and x=1.

For $\int_{-2}^{1} f(x)\,dx$, the video explains that one can compute the ordinary geometric area of the region below the axis and then attach a negative sign because the function lies below the x-axis.

1. The interval is traversed from lower limit to higher limit.

2. The function is below the x-axis throughout the region being measured.

3. For a Riemann-integrable function on an ordered finite interval a<b, the definite integral is net signed area: regions above the axis add, while regions below subtract. The displayed piecewise curve is continuous, which is sufficient for integrability. The curved pieces are the semicircles specified in this geometric exercise, rather than arbitrary curves inferred to be circular.

1. Definite integral as geometric area above the x-axis

1. Audio
   Observation

   The lower trapezoid can be measured directly or decomposed into simpler polygonal pieces.

2. Diagram
   Observation

   The lower region is visually decomposed into simpler polygonal pieces.

The pictured lower quadrilateral is a trapezoid. Its geometric area can be found as one trapezoid or as a rectangle plus two triangles; this is not a rule that every polygon is a trapezoid.

1. The region has straight-line boundaries so it can be treated as a polygon.

1. Definite integral is negative when the graph is below the x-axis

1. Audio
   Observation

   The explanation starts measuring the triangular component from its base and height.

2. Diagram
   Observation

   One triangular piece of the lower region is singled out during the decomposition.

The triangular piece has base1 and height2, so its geometric area is1. A unit-area label is already visible near the end of this initial section.

1. The chosen piece is a triangle with base 1 and height 2.

1. Computing polygonal area by trapezoid formula or decomposition

1. Audio
   Observation

   The worked regions illustrate positive contributions above the axis and negative contributions below it.

2. Diagram
   Observation

   Shaded regions are drawn between the graph and the x-axis for each requested interval.

For a Riemann-integrable function on an ordered finite interval a<b, the definite integral is net signed area: regions above the axis add, while regions below subtract. The displayed piecewise curve is continuous, which is sufficient for integrability. The curved pieces are the semicircles specified in this geometric exercise, rather than arbitrary curves inferred to be circular.

1. The graph of $f$ is available on $[a,b]$.

2. The region can be decomposed into standard shapes whose areas are known.

3. A single sign applies only when that region is on one side of the axis; otherwise add its positive and negative pieces separately.

1. Audio
   Observation

   The triangular component is measured by one-half times its base times perpendicular height.

2. Formula
   Observation

   The local small-triangle computation uses one-half times base1 times height2, giving area1; the larger triangle is treated later.

The area of a triangle equals one-half times its base times its height. The video applies this twice: once to a small triangle of base 1 and height 2, and once to a larger triangle of base 3 and height 4.

1. $b$ is a chosen base length.

2. $h$ is the perpendicular height corresponding to that base.

1. Audio
   Observation

   The rectangle component is measured by multiplying its two side lengths.

2. Diagram
   Observation

   A rectangular portion of the shaded region below the x-axis is identified on the graph.

The area of a rectangle is length times width. In the worked example, the rectangle has side lengths 2 and 1, giving area 2.

1. $l$ and $w$ are the side lengths of the rectangle.

1. Audio
   Observation

   The circle-area formula is applied with the final semicircle radius, and the full-circle area is halved.

2. Formula
   Observation

   Written as $\pi \cdot 1^2$, then divided by 2 for a semicircle.

The area of a full circle is $\pi r^2$. If only half of the circle is present, as in the final interval, the area is halved: $\frac{\pi r^2}{2}$. With $r=1$, the semicircle area is $\frac{\pi}{2}$.

1. The region is a full circle or exactly half of a circle.

2. $r$ is the radius.

1. Audio
   Observation

   The polygon pieces add to the ordinary area, then the lower region gives the negative integral result.

2. Formula
   Observation

   The answer for $\int_{-2}^{1} f(x)\,dx$ is written as $-4$.

If the relevant region between the graph and the x-axis lies below the x-axis, then the definite integral over that interval equals the negative of the ordinary geometric area.

1. The graph of $f$ is below the x-axis on the interval under consideration.

2. The total unsigned area of the region has been computed.

3. For a Riemann-integrable function on an ordered finite interval a<b, the definite integral is net signed area: regions above the axis add, while regions below subtract. The displayed piecewise curve is continuous, which is sufficient for integrability. The curved pieces are the semicircles specified in this geometric exercise, rather than arbitrary curves inferred to be circular.

For the shown ordered intervals with nonpositive f and positive ordinary area; endpoints on the axis contribute zero.

1. Audio
   Observation

   The upper triangle uses the shown base and height, so its positive geometric area gives the integral.

2. Formula
   Observation

   Written result: $\frac12 \cdot 3 \cdot 4 = 6$.

When the bounded region lies above the x-axis, the definite integral equals the positive geometric area of that region.

1. The graph of $f$ is above the x-axis on the interval under consideration.

2. The region’s geometric area has been computed.

3. For a Riemann-integrable function on an ordered finite interval a<b, the definite integral is net signed area: regions above the axis add, while regions below subtract. The displayed piecewise curve is continuous, which is sufficient for integrability. The curved pieces are the semicircles specified in this geometric exercise, rather than arbitrary curves inferred to be circular.

For the shown ordered interval with nonnegative f and positive ordinary area.

1. Audio
   Observation

   The first calculation halves the circle area and reaches the displayed semicircle result.

2. Formula
   Observation

   The handwritten chain shows (π·2²)/2 = 2π.

3. Diagram
   Observation

   The shaded semicircular region from x=-6 to x=-2 is identified as the relevant area.

1. Expression

   $$\int_{-6}^{-2} f(x)\,dx$$
   Explanation

   Start with the first requested definite integral.

   Justification

   This is the problem statement shown on screen.

   Shown in the video
2. Expression

   $$\text{area under } f \text{ and above the } x\text{-axis on }[-6,-2]$$
   Explanation

   Interpret the integral as the geometric area of the shaded region.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
3. Expression

   $$\text{region is a semicircle with radius }2$$
   Explanation

   Identify the shape and read off its radius from the graph.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
4. Expression

   $$\frac{\pi r^2}{2}=\frac{\pi\cdot 2^2}{2}$$
   Explanation

   Use the circle area formula and take half because the region is a semicircle.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
5. Expression

   $$2\pi$$
   Explanation

   Simplify the expression to obtain the value of the integral.

   Justification

   Arithmetic simplification of (π·2²)/2.

   Shown in the video

$\int_{-6}^{-2} f(x)\,dx$ = 2π.

1. Audio
   Observation

   The second calculation establishes the negative sign and begins the decomposition of the lower region.

2. Diagram
   Observation

   The region from x=-2 to x=1 is shaded below the axis and then split into simpler shapes.

3. Formula
   Observation

   One triangle is written as 1 times 2 times one-half.

1. At this early timestamp the second calculation is still in progress; later in the same complete video it reaches −4 and the final two problems are also solved.

1. Expression

   $$\int_{-2}^{1} f(x)\,dx$$
   Explanation

   Begin the second requested definite integral.

   Justification

   This is the next problem shown and spoken aloud.

   Shown in the video
2. Expression

   $$f(x)\le0\quad(-2\le x\le1)$$
   Explanation

   The graph is nonpositive on the closed interval and negative in its interior; both endpoints are on the axis.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Supplementary explanation
3. Expression

   $$\int_{-2}^{1} f(x)\,dx = -(\text{geometric area of the region})$$
   Explanation

   Convert the geometric area into a signed integral value by adding a minus sign.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
4. Expression

   $$\text{area can be found as a trapezoid or as rectangle + two triangles}$$
   Explanation

   Choose a method for computing the geometric area of the polygonal region.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
5. Expression

   $$A_{\triangle}=\frac{1\cdot 2}{2}$$
   Explanation

   Compute one triangular piece in the decomposition.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video

At this early timestamp the second calculation is still in progress; later in the same complete video it reaches −4 and the final two problems are also solved.

1. Audio
   Observation

   The polygon pieces add to the ordinary area, then the lower region gives the negative integral result.

2. Formula
   Observation

   Final written answer is $-4$.

1. Expression

   $$A_1=1$$
   Explanation

   A small triangular piece of the shaded region has area 1.

   Justification

   Triangle area formula $\frac12 bh$ with base 1 and height 2.

   Shown in the video
2. Expression

   $$A_2=2$$
   Explanation

   A rectangular piece has area 2.

   Justification

   Rectangle area formula $lw$ with side lengths 2 and 1.

   Shown in the video
3. Expression

   $$A_3=1$$
   Explanation

   Another triangular piece also has area 1.

   Justification

   Triangle area formula $\frac12 bh$ with base 1 and height 2.

   Shown in the video
4. Expression

   $$A_{\text{total}}=1+2+1=4$$
   Explanation

   Add the three unsigned areas to get the total geometric area.

   Justification

   Additivity of area for non-overlapping pieces.

   Shown in the video
5. Expression

   $$\int_{-2}^{1} f(x)\,dx=-4$$
   Explanation

   Because the whole region lies below the x-axis, the signed integral is the negative of the total area.

   Justification

   Signed-area interpretation of the definite integral.

   Shown in the video

$\int_{-2}^{1} f(x)\,dx=-4$.

1. Audio
   Observation

   The upper triangle uses the shown base and height, so its positive geometric area gives the integral.

2. Formula
   Observation

   Written computation: $\frac12 \cdot 3 \cdot 4 = 6$.

1. Expression

   $$b=4-1=3$$
   Explanation

   The base of the triangle spans from $x=1$ to $x=4$.

   Justification

   Reading the horizontal extent from the graph.

   Shown in the video
2. Expression

   $$h=4$$
   Explanation

   The height of the triangle is 4 units.

   Justification

   Reading the vertical extent from the graph.

   Shown in the video
3. Expression

   $$A=\frac12 \cdot 3 \cdot 4=6$$
   Explanation

   Apply the triangle area formula.

   Justification

   $A_{\triangle}=\frac12 bh$.

   Shown in the video
4. Expression

   $$\int_{1}^{4} f(x)\,dx=6$$
   Explanation

   The region is above the x-axis, so the signed area is positive.

   Justification

   Signed-area interpretation of the definite integral.

   Shown in the video

$\int_{1}^{4} f(x)\,dx=6$.

1. Audio
   Observation

   The lower curved region is identified as a semicircle, its area is halved from the circle formula, and its integral receives a negative sign.

2. Formula
   Observation

   Written result: $-\frac{\pi \cdot 1^2}{2}=-\frac{\pi}{2}$.

1. Expression

   $$r=1$$
   Explanation

   The curved region is identified as a semicircle of radius 1.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
2. Expression

   $$A_{\text{circle}}=\pi r^2=\pi\cdot 1^2$$
   Explanation

   Start from the full-circle area formula.

   Justification

   Circle area formula.

   Shown in the video
3. Expression

   $$A_{\text{semicircle}}=\frac{\pi\cdot 1^2}{2}$$
   Explanation

   Only half of the circle is present, so divide by 2.

   Justification

   Semicircle is half of a circle.

   Shown in the video
4. Expression

   $$\int_{4}^{6} f(x)\,dx=-\frac{\pi\cdot 1^2}{2}=-\frac{\pi}{2}$$
   Explanation

   The semicircular region lies below the x-axis, so the integral is negative.

   Justification

   Signed-area interpretation of the definite integral.

   Shown in the video

$\int_{4}^{6} f(x)\,dx=-\frac{\pi}{2}$.

1. Formula
   Observation

   The first listed integral is $\int_{-6}^{-2} f(x)\,dx$ = ?.

2. Audio
   Observation

   The first calculation halves the circle area and reaches the displayed semicircle result.

3. Diagram
   Observation

   The semicircular region above the x-axis is shaded.

Find the definite integral $\int_{-6}^{-2} f(x)\,dx$ from the given graph of f.

1. The graph of f is provided.

2. On [-6,-2], the graph forms a semicircle above the x-axis.

3. The semicircle has radius 2.

Compute the exact value of the definite integral.

1. Expression

   $$\int_{-6}^{-2} f(x)\,dx$$
   Explanation

   Identify the target integral.

   Justification

   Given in the problem list on screen.

   Shown in the video
2. Expression

   $$\text{area of shaded region above the }x\text{-axis}$$
   Explanation

   Translate the integral into an area problem.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
3. Expression

   $$\frac{\pi\cdot 2^2}{2}$$
   Explanation

   Use the semicircle area formula with radius 2.

   Justification

   The region is identified as a semicircle and the circle area formula is πr².

   Shown in the video
4. Expression

   $$2\pi$$
   Explanation

   Simplify to the final value.

   Justification

   Direct arithmetic simplification.

   Shown in the video

$2\pi$

The result matches the handwritten final expression shown beside the first integral.

1. Formula
   Observation

   The second listed integral is $\int_{-2}^{1} f(x)\,dx$ = ?.

2. Audio
   Observation

   The second calculation establishes the negative sign and begins the decomposition of the lower region.

3. Diagram
   Observation

   The region below the x-axis is shaded and split into simpler polygonal parts.

1. At this early timestamp the second calculation is still in progress; later in the same complete video it reaches −4 and the final two problems are also solved.

Find the definite integral $\int_{-2}^{1} f(x)\,dx$ from the same graph of f.

1. The graph of f is provided.

2. On [-2,1], the relevant region lies below the x-axis.

3. The lower region can be viewed as a trapezoid or decomposed into a rectangle and two triangles.

4. One triangle has base 1 and height 2.

Compute the definite integral using geometric area and sign.

1. Expression

   $$\int_{-2}^{1} f(x)\,dx$$
   Explanation

   Identify the second target integral.

   Justification

   Given in the problem list on screen.

   Shown in the video
2. Expression

   $$\text{function is below the }x\text{-axis}$$
   Explanation

   Recognize that the region contributes negative signed area.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
3. Expression

   $$-(\text{geometric area})$$
   Explanation

   Plan to compute the ordinary area first and then negate it.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
4. Expression

   $$\text{trapezoid or rectangle + two triangles}$$
   Explanation

   Choose a geometric decomposition for the area.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
5. Expression

   $$\frac{1\cdot 2}{2}$$
   Explanation

   Begin computing one triangular component.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video

At this early timestamp the second calculation is still in progress; later in the same complete video it reaches −4 and the final two problems are also solved.

The later same-video result −4 confirms the negative sign and ordinary area4.

1. Formula
   Observation

   Prompt on screen: $\int_{-2}^{1} f(x)\,dx=?$; final written answer $-4$.

2. Audio
   Observation

   The polygon pieces add to the ordinary area, then the lower region gives the negative integral result.

Find $\int_{-2}^{1} f(x)\,dx$ from the given graph.

1. Graph of $f$ on the coordinate plane.

2. Interval $[-2,1]$.

3. Region below the x-axis composed of two triangles and one rectangle.

Compute the definite integral as signed area.

1. Expression

   $$A_{\triangle}=\frac12\cdot 1\cdot 2=1$$
   Explanation

   Compute the first small triangle’s area.

   Justification

   Triangle area formula.

   Shown in the video
2. Expression

   $$A_{\text{rect}}=2\cdot 1=2$$
   Explanation

   Compute the rectangle’s area.

   Justification

   Rectangle area formula.

   Shown in the video
3. Expression

   $$A_{\triangle}=\frac12\cdot 1\cdot 2=1$$
   Explanation

   Compute the second small triangle’s area.

   Justification

   Triangle area formula.

   Shown in the video
4. Expression

   $$1+2+1=4$$
   Explanation

   Add the unsigned areas.

   Justification

   Total area is the sum of the pieces.

   Shown in the video
5. Expression

   $$\int_{-2}^{1} f(x)\,dx=-4$$
   Explanation

   Attach a negative sign because the region is below the x-axis.

   Justification

   Signed-area rule for definite integrals.

   Shown in the video

$-4$

Matches the written final answer on screen.

1. Formula
   Observation

   Prompt on screen: $\int_{1}^{4} f(x)\,dx=?$; written computation $\frac12\cdot 3\cdot 4=6$.

2. Audio
   Observation

   The upper triangle uses the shown base and height, so its positive geometric area gives the integral.

Find $\int_{1}^{4} f(x)\,dx$ from the given graph.

1. Graph of $f$.

2. Interval $[1,4]$.

3. Region above the x-axis shaped like a triangle.

Compute the definite integral as signed area.

1. Expression

   $$b=3$$
   Explanation

   The base runs from $x=1$ to $x=4$.

   Justification

   Horizontal distance on the graph.

   Shown in the video
2. Expression

   $$h=4$$
   Explanation

   The height is 4 units.

   Justification

   Vertical extent read from the graph.

   Shown in the video
3. Expression

   $$\frac12\cdot 3\cdot 4=6$$
   Explanation

   Apply the triangle area formula.

   Justification

   $A_{\triangle}=\frac12 bh$.

   Shown in the video
4. Expression

   $$\int_{1}^{4} f(x)\,dx=6$$
   Explanation

   The region is above the x-axis, so the signed area is positive.

   Justification

   Signed-area interpretation.

   Shown in the video

$6$

Matches the written final answer on screen.

1. Formula
   Observation

   Prompt on screen: $\int_{4}^{6} f(x)\,dx=?$; written result $-\frac{\pi\cdot 1^2}{2}=-\frac{\pi}{2}$.

2. Audio
   Observation

   The lower curved region is identified as a semicircle, its area is halved from the circle formula, and its integral receives a negative sign.

Find $\int_{4}^{6} f(x)\,dx$ from the given graph.

1. Graph of $f$.

2. Interval $[4,6]$.

3. Region below the x-axis shaped like a semicircle of radius 1.

Compute the definite integral as signed area.

1. Expression

   $$r=1$$
   Explanation

   Identify the radius of the semicircle.

   Justification

   Uses the displayed region, its recognized shape and the area formula explained at this stage.

   Shown in the video
2. Expression

   $$A_{\text{full circle}}=\pi r^2=\pi\cdot 1^2$$
   Explanation

   Write the full-circle area.

   Justification

   Circle area formula.

   Shown in the video
3. Expression

   $$A_{\text{semicircle}}=\frac{\pi\cdot 1^2}{2}$$
   Explanation

   Take half because only a semicircle appears.

   Justification

   Semicircle is half a circle.

   Shown in the video
4. Expression

   $$\int_{4}^{6} f(x)\,dx=-\frac{\pi\cdot 1^2}{2}=-\frac{\pi}{2}$$
   Explanation

   Negate because the region is below the x-axis.

   Justification

   Signed-area interpretation.

   Shown in the video

$-\frac{\pi}{2}$

Matches the written final answer on screen.

1. Diagram
   Observation

   A coordinate grid with labeled axes and a teal graph of f is visible throughout.

2. Formula
   Observation

   Four integrals are listed on the left side of the screen.

1. Cartesian coordinate plane

2. Graph of f

3. List of four definite integrals

4. Purple pointer dot

1. The pointer moves among the listed integrals and corresponding parts of the graph.

2. Handwritten annotations appear next to the first two integrals as they are discussed.

1. The graph of f remains fixed on the right.

2. The four integral prompts remain listed on the left.

The visual setup ties each algebraic integral to a specific region under the graph of f.

1. Diagram
   Observation

   The region from x=-6 to x=-2 above the x-axis is shaded.

2. Formula
   Observation

   The handwritten expression (π·2²)/2 = 2π appears beside the first integral.

1. Semicircular region above the x-axis

2. Radius marks labeled 2

3. Handwritten area formula

1. The semicircular region becomes shaded.

2. Radius annotations and the area computation are written in.

1. The rest of the graph stays unchanged.

The animation visually converts the first definite integral into a semicircle area problem.

1. Diagram
   Observation

   The region from x=-2 to x=1 below the x-axis is shaded.

2. Animation
   Observation

   The lower region is then subdivided into simpler polygonal pieces.

1. At this early timestamp the second calculation is still in progress; later in the same complete video it reaches −4 and the final two problems are also solved.

1. Polygonal region below the x-axis

2. Rectangle piece

3. Two triangular pieces

4. Trapezoid interpretation

1. The lower region is shaded.

2. Internal lines divide the region into simpler shapes for area calculation.

1. The outer boundary of the region from x=-2 to x=1 remains the same.

The visual process supports the spoken method of computing geometric area first and then assigning a negative sign to the integral.

1. Diagram
   Observation

   The graph shows several colored regions between the curve and the x-axis, corresponding to the four requested integrals.

2. Animation
   Observation

   As each integral is discussed, the relevant region is visually emphasized.

1. Exact color coding of every subregion is not fully described in the audio.

1. Coordinate axes labeled $x$ and $y$.

2. Graph of $f$.

3. Shaded regions over $[-6,-2]$, $[-2,1]$, $[1,4]$, and $[4,6]$.

4. List of four definite integrals on the left.

1. Attention shifts from one integral to the next.

2. Each target region is highlighted when its integral is discussed.

1. The underlying graph of $f$ remains the same throughout.

2. The four requested integrals remain listed on screen.

The visual setup supports interpreting each definite integral as the signed area between the graph and the x-axis over the specified interval.

1. Diagram
   Observation

   The region for $[-2,1]$ is broken into simpler geometric pieces.

2. Audio
   Observation

   The polygon pieces add to the ordinary area, then the lower region gives the negative integral result.

1. One small triangle.

2. One rectangle.

3. Another small triangle.

1. The speaker moves piece by piece through the composite region.

2. The separate areas are summed into one total.

1. All pieces lie below the x-axis.

2. The overall interval remains $[-2,1]$.

A composite region can be evaluated by splitting it into standard shapes, summing their areas, and then applying the correct sign.

1. Diagram
   Observation

   The region for $[1,4]$ is a single triangle above the x-axis.

2. Audio
   Observation

   The upper triangle uses the shown base and height, so its positive geometric area gives the integral.

1. One triangle bounded by the graph and the x-axis.

1. Base and height are identified directly from the graph.

1. The region stays entirely above the x-axis.

When the region is a single simple shape, its area can be read off directly and used as the positive value of the integral.

1. Diagram
   Observation

   The region for $[4,6]$ is a semicircular cap below the x-axis.

2. Audio
   Observation

   The lower curved region is identified as a semicircle, its area is halved from the circle formula, and its integral receives a negative sign.

1. One semicircle of radius 1.

1. Full-circle area formula is adapted to a semicircle by dividing by 2.

2. Sign is then reversed because the region is below the axis.

1. The shape remains circular/semicircular throughout the explanation.

Curved regions can still be handled by recognizing them as fractions of circles and applying the same signed-area principle.

1. Audio
   Observation

   The explanation identifies the lower region and assigns a negative sign to its integral contribution.

One might compute only the geometric area of a region below the x-axis and forget that the definite integral carries a negative sign.

The video explicitly warns that when the function is below the x-axis, the integral equals the negative of the geometric area.

1. Audio
   Observation

   The polygon pieces add to the ordinary area, then the lower region gives the negative integral result.

After adding the areas of the pieces, one may incorrectly conclude that $\int_{-2}^{1} f(x)\,dx=4$.

The definite integral is signed area. Since the region lies below the x-axis, the correct value is $-4$, not $4$.

1. Audio
   Observation

   The first positive region is linked to the definite integral by its geometric area.

The definition of the definite integral as area is applied to a semicircular region in the first example.

1. Audio
   Observation

   The lower trapezoid can be measured directly or decomposed into simpler polygonal pieces.

The signed-area rule determines the final sign, while the trapezoid/decomposition method supplies the geometric area magnitude.

1. Audio
   Observation

   The explanation starts measuring the triangular component from its base and height.

The decomposition method contains the triangle-area computation as one of its substeps.

1. Audio
   Observation

   The triangular component is measured by one-half times its base times perpendicular height.

The signed-area method reduces the integral to geometric area computations, including the triangle formula.

1. Audio
   Observation

   The rectangle component is measured by multiplying its two side lengths.

The same integral-evaluation method uses the rectangle area formula for one subregion.

1. Audio
   Observation

   The circle-area formula is applied with the final semicircle radius, and the full-circle area is halved.

For the last interval, the signed-area method relies on the circle/semicircle area formula.

1. Audio
   Observation

   The worked regions illustrate positive contributions above the axis and negative contributions below it.

Regions below the x-axis contribute negative signed area, while regions above contribute positive signed area.

1. Audio
   Observation

   The upper curved region is identified as a semicircle with its shown radius.

1. Definite integral as geometric area above the x-axis
2. Area of a semicircle from the circle area formula
3. Evaluation of $\int_{-6}^{-2} f(x)\,dx$ by semicircle area
4. Example 1: $\int_{-6}^{-2} f(x)\,dx$

1. Audio
   Observation

   The explanation identifies the lower region and assigns a negative sign to its integral contribution.

1. Definite integral is negative when the graph is below the x-axis
2. Do not treat area below the x-axis as positive in the definite integral
3. Partial evaluation of $\int_{-2}^{1} f(x)\,dx$ using signed area and decomposition

1. Audio
   Observation

   The lower trapezoid can be measured directly or decomposed into simpler polygonal pieces.

1. Computing polygonal area by trapezoid formula or decomposition
2. Triangle area computation inside the second example
3. Partial evaluation of $\int_{-2}^{1} f(x)\,dx$ using signed area and decomposition

1. Diagram
   Observation

   The screen pairs listed integrals with shaded regions on the graph of f.

1. Definite integral as geometric area above the x-axis
2. Overall layout of the worked problem
3. Shading and annotation for the first integral
4. Shading and decomposition for the second integral

1. Audio
   Observation

   The polygon pieces add to the ordinary area, then the lower region gives the negative integral result.

1. Evaluate definite integrals from a graph using signed area
2. Integral over a region below the x-axis is negative
3. Mistaking total geometric area for the definite integral

1. Audio
   Observation

   The upper triangle uses the shown base and height, so its positive geometric area gives the integral.

1. Area formula for a triangle
2. Example: $\int_{1}^{4} f(x)\,dx$

1. Audio
   Observation

   The lower curved region is identified as a semicircle, its area is halved from the circle formula, and its integral receives a negative sign.

1. Area formula for a circle and semicircle
2. Example: $\int_{4}^{6} f(x)\,dx$

1. Audio
   Observation

   The worked regions illustrate positive contributions above the axis and negative contributions below it.

1. Evaluate definite integrals from a graph using signed area