AP Calculus AB Unit 8: Applications of Integration — Free Review Games.

Area \(= \int_0^2 x^2 \, dx = \frac{x^3}{3} \Big|_0^2 = 8/3\).

The area between curves is the integral of the top function minus the bottom function over the interval.

When velocity is non-negative, the integral equals displacement, which is the same as total distance traveled.

The disk method: V = pi * integral from a to b of [f(x)]^2 dx. Each disk has radius f(x) and thickness dx.

The accumulation function gives the net accumulated value (area with sign) of f from the starting point to x.

Area \(= \int_0^1 (x - x^2) \, dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = 1/2 - 1/3 = 1/6\).

\(V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi \cdot 8 = 8\pi\).

v = 0 at t = 3. Distance = |integral 0 to 3 of (t-3) dt| + integral 3 to 5 of (t-3) dt = |-9/2| + 2 = 4.5 + 2 = 6.5.

\(V = \pi \int_0^1 [(x)^2 - (x^2)^2] \, dx = \pi \int [x^2 - x^4] \, dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi(1/3 - 1/5) = 2\pi/15\).

Displacement = integral from 0 to pi of cos(t) dt = sin(t) from 0 to pi = sin(pi) - sin(0) = 0 - 0 = 0.

Area = integral from 0 to pi of sin(x) dx + |integral from pi to 2pi of sin(x) dx| = 2 + |-(-2)| = 2 + 2 = 4.

Side of square \(= \sqrt{x}\). Area of cross-section \(= (\sqrt{x})^2 = x\). \(V = \int_0^4 x \, dx = \frac{x^2}{2} \Big|_0^4 = 8\).

Shell method: \(V = 2\pi \int_0^2 x(4 - x^2) \, dx = 2\pi \left[2x^2 - \frac{x^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi\).

Net change = integral from 0 to 2 of (3t^2 - t) dt = [t^3 - t^2/2] from 0 to 2 = (8 - 2) - 0 = 6 gallons.

Area = integral from 0 to ln(2) of (e^x - e^(-x)) dx = [e^x + e^(-x)] from 0 to ln(2) = (2 + 1/2) - (1 + 1) = 5/2 - 2 = 1/2.

Subtracting the lower function \(g(x)\) from the upper function \(f(x)\) guarantees the integrand is nonnegative, so the integral gives the vertical strip height times \(dx\) summed correctly. The choice \(\int_a^b [g(x)-f(x)]\,dx\) is wrong because it would produce a negative value since \(g(x) \leq f(x)\) on this interval. Always identify which curve is on top before subtracting to avoid computing a negative or meaningless area.

Each cross-sectional slice is a circular disk with radius \(f(x)\), so its area is \(\pi [f(x)]^2\), and integrating this area over \([a,b]\) gives the volume. The expression \(2\pi\int_a^b x f(x)\,dx\) is instead the shell method formula for rotation about the y-axis, not the disk method about the x-axis. Remember that disk/washer volumes always include the factor \(\pi\) times the square of the radius function.

The washer's area is the outer disk area minus the inner disk area, \(\pi R(x)^2 - \pi r(x)^2\), so squaring each radius separately before subtracting is required. The choice \(\pi\int_a^b [R(x)-r(x)]^2\,dx\) is incorrect because squaring the difference of radii does not equal the difference of squared radii. Always square each radius individually in washer problems rather than squaring the combined expression.

Each cylindrical shell has circumference \(2\pi x\), height \(f(x)\), and infinitesimal thickness \(dx\), so its volume is \(2\pi x f(x)\,dx\), which integrates to give the total volume. The option \(\pi\int_a^b [f(x)]^2\,dx\) is instead a disk-method formula for rotation about the x-axis, not a shell formula for rotation about the y-axis. The shell method is most useful when the axis of rotation is vertical and the region is more naturally described as a function of \(x\).

The Fundamental Theorem of Calculus, Part 1, states that differentiating an accumulation function with respect to its upper limit simply returns the integrand evaluated at that limit, giving \(f(x)\). The choice \(f'(x)\) is wrong because it would require differentiating \(f\) itself, which is not what the theorem does. This rule lets you find derivatives of accumulation functions instantly without evaluating the integral.

Subtracting the lower curve from the upper curve ensures each vertical strip's height is a positive quantity, which is essential since area must be a nonnegative measurement. The option about convergence is irrelevant here because a continuous function on a closed interval is automatically integrable regardless of which curve is subtracted. Always check which function is greater on the interval in question before setting up the subtraction to avoid a negative or incorrect area.

A square cross-section has area \([s(x)]^2\), so integrating this area function over \([a,b]\) using the standard cross-section method gives the total volume, no \(\pi\) factor needed since squares are not circles. The choice \(\pi\int_a^b [s(x)]^2\,dx\) incorrectly applies the circular disk formula to a square cross-section. Always match the cross-sectional area formula to the actual shape described in the problem, whether it is a square, semicircle, or equilateral triangle.

An integral with identical upper and lower limits represents zero width, so there is no area to accumulate, making the value exactly \(0\) regardless of the function \(f\). The choice \(f(a)\) is incorrect because that would be the value of the integrand, not the accumulated area. This property is useful for evaluating accumulation functions at their starting point.

Integrating velocity directly gives the net change in position, or net displacement, because positive and negative contributions can cancel when the particle changes direction. The choice "Total distance traveled" is incorrect because that requires integrating \(|v(t)|\) instead, which accounts for direction changes separately. Distinguishing net displacement from total distance is a critical AP Calculus skill whenever velocity changes sign.

Taking the absolute value of velocity before integrating converts negative contributions (motion in the reverse direction) into positive ones, so the sum represents the total path length traveled rather than the net position change. The choice "Net displacement" is wrong because that is obtained from \(\int_a^b v(t)\,dt\) without the absolute value. When velocity changes sign on the interval, distance traveled will always be greater than or equal to the magnitude of net displacement.

When integrating with respect to \(y\), horizontal strips have width \(g(y)\) and thickness \(dy\), so the area is simply \(\int_c^d g(y)\,dy\), analogous to the x-axis case. The option \(\pi\int_c^d [g(y)]^2\,dy\) is instead a volume formula for a solid of revolution, not an area calculation. Recognize when a region is more naturally integrated with respect to \(y\), such as when curves are given as \(x\) in terms of \(y\).

The radius of the disk is the vertical distance from the axis of rotation \(y=k\) up to the curve \(f(x)\), which is \(f(x)-k\) since \(f(x) \ge k\). The choice \(k-f(x)\) is wrong because it would give a negative radius under the stated condition, which is not physically meaningful. Whenever rotating about a horizontal line other than the x-axis, always adjust the radius by subtracting the axis's y-value.

Setting the two functions equal and solving locates the x-values where the curves cross, which become the natural bounds of integration for the enclosed region. Taking derivatives is unrelated to this step since derivatives describe slopes, not intersection locations. Always solve for intersection points first before setting up any area or volume integral involving two curves.

Setting \(x^2=2x\) gives intersections at \(x=0\) and \(x=2\), and since \(2x \ge x^2\) on this interval, the area is \(\int_0^2 (2x-x^2)\,dx = [x^2-\frac{x^3}{3}]_0^2 = 4-\frac{8}{3}=\frac{4}{3}\). The choice \(\frac{8}{3}\) is wrong because it comes from forgetting to subtract the \(x^2\) term's contribution correctly. Always verify which curve lies above the other over the entire interval before integrating the difference.

Using the disk method, the volume is \(\pi\int_0^2 (x^3)^2\,dx = \pi\int_0^2 x^6\,dx = \pi\left[\frac{x^7}{7}\right]_0^2 = \frac{128\pi}{7}\). The choice \(\frac{64\pi}{7}\) results from an arithmetic error in evaluating \(2^7\) rather than correctly computing \(128\). When applying the disk method, remember the radius must be squared before integrating, giving a power one higher than expected.

Evaluating the antiderivative gives \(F(x)=t^2+t\) evaluated from \(0\) to \(x\), so \(F(3)=9+3=12\). The choice \(9\) is incorrect because it omits the linear term \(+t\) from the antiderivative. When evaluating an accumulation function at a specific value, always find the full antiderivative before substituting the bounds.

Since \(v(t)=3t(t-2)\) is negative on \((0,2)\) and positive on \((2,3)\), the total distance is \(|\int_0^2 v\,dt| + \int_2^3 v\,dt = |-4| + 4 = 8\). The choice \(0\) is wrong because it represents the net displacement, where the negative and positive portions cancel, not the total distance traveled. Whenever velocity changes sign, split the interval at the sign change and take absolute values before summing.

Since \(\cos x \ge \sin x\) on \([0,\pi/4]\), the area is \(\int_0^{\pi/4} (\cos x - \sin x)\,dx = [\sin x + \cos x]_0^{\pi/4} = \sqrt{2} - 1\). The choice \(1-\sqrt{2}\) is wrong because it reverses the order of the curves, giving a negative area. Confirm which trigonometric function dominates over the interval before subtracting, since sine and cosine frequently switch dominance at multiples of \(\pi/4\).

The shell volume is \(2\pi\int_0^2 x \cdot x^2\,dx = 2\pi\int_0^2 x^3\,dx = 2\pi\left[\frac{x^4}{4}\right]_0^2 = 2\pi(4) = 8\pi\). The choice \(16\pi\) comes from forgetting to divide by 4 when evaluating the antiderivative \(\frac{x^4}{4}\). Remember that the shell method radius is \(x\) and height is the function value when rotating about the y-axis.

By the Fundamental Theorem of Calculus, differentiating an accumulation function returns the integrand evaluated at \(x\), giving \(F'(x)=\frac{1}{x}\). The choice \(\ln x\) is incorrect because that is the value of \(F(x)\) itself, not its derivative. Do not confuse the accumulation function with its derivative when applying FTC Part 1.

Since \(x^2-4 \le 0\) on \([-2,2]\), the actual area is the absolute value of \(\int_{-2}^2 (x^2-4)\,dx = -\frac{32}{3}\), giving an area of \(\frac{32}{3}\). The choice \(\frac{16}{3}\) results from a computational error when evaluating the antiderivative at the bounds. When a curve dips below the x-axis, remember that the raw integral is negative and must be converted to a positive area value.

Using the disk method, \(V=\pi\int_0^1 (e^x)^2\,dx = \pi\int_0^1 e^{2x}\,dx = \frac{\pi}{2}[e^{2x}]_0^1 = \frac{\pi}{2}(e^2-1)\). The choice \(\pi(e^2-1)\) omits the factor of \(\frac{1}{2}\) that arises from integrating \(e^{2x}\). Always account for the chain-rule factor when the exponent involves a coefficient other than 1.

The average value formula is \(\frac{1}{b-a}\int_a^b f(x)\,dx = \frac{12}{5-2}=4\). The choice \(12\) is incorrect because it is simply the value of the integral without dividing by the interval length. Always divide the accumulated total by the width of the interval to find an average value.

Since \(x^3-x\) is negative on \((0,1)\), the area is \(\left|\int_0^1 (x^3-x)\,dx\right| = \left|\frac{1}{4}-\frac{1}{2}\right| = \frac{1}{4}\). The choice \(0\) is wrong because it incorrectly assumes the curve does not deviate from the axis, ignoring that the raw integral is nonzero even though it is negative. Take the absolute value of a negative definite integral whenever it is being interpreted as a geometric area.

Since \(x \ge x^2\) on \([0,1]\), the outer radius is \(x\) and the inner radius is \(x^2\), giving \(V=\pi\int_0^1 (x^2-x^4)\,dx = \pi\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{2\pi}{15}\). The choice \(\frac{\pi}{3}\) mistakenly uses only the outer radius term \(x^2\) without subtracting the inner radius term \(x^4\). Always square both the outer and inner radius separately before subtracting in a washer setup.

The displacement equals \(\int_2^6 (4-\tau)\,d\tau = [4\tau - \frac{\tau^2}{2}]_2^6 = 6 - 6 = 0\), meaning the particle returns to its starting position on this subinterval. The choice \(12\) is wrong because it fails to account for the negative velocity contribution as \(\tau\) exceeds 4. Remember that \(s(b)-s(a)\) equals the definite integral of velocity over \([a,b]\), regardless of sign changes within the interval.

Since \(\sqrt{x} \ge x\) on \([0,1]\), the outer radius squared is \((\sqrt{x})^2=x\) and the inner radius squared is \(x^2\), giving \(\pi\int_0^1 (x-x^2)\,dx\). The choice \(\pi\int_0^1 (\sqrt{x}-x)^2\,dx\) is wrong because it squares the difference of the radii instead of subtracting the squares of the radii individually. Never square the combined expression \((R-r)\) when applying the washer formula; always square \(R\) and \(r\) separately first.

Evaluating the antiderivative gives \(g(x)=2x-\frac{x^2}{2}\), so \(g(4)=8-8=0\), since the positive contribution from \(t \in (0,2)\) exactly cancels the negative contribution from \(t \in (2,4)\). The choice \(8\) is incorrect because it stops after the first term of the antiderivative and ignores the subtraction of \(\frac{x^2}{2}\). When \(f\) changes sign, the accumulation function can return to zero even over a nonzero interval.

Setting \(x^2=-x^2+8\) gives \(x=\pm2\), and since \(-x^2+8 \ge x^2\) on \([-2,2]\), the area is \(\int_{-2}^2 (8-2x^2)\,dx = \left[8x-\frac{2x^3}{3}\right]_{-2}^2 = \frac{64}{3}\). The choice \(32\) is wrong because it comes from omitting the \(x^3\) term's contribution when evaluating the antiderivative. Always simplify the combined integrand \((f-g)\) fully before integrating symmetric regions like this one.

Using the disk method, \(V=\pi\int_1^3 \frac{1}{x^2}\,dx = \pi\left[-\frac{1}{x}\right]_1^3 = \pi\left(1-\frac{1}{3}\right)=\frac{2\pi}{3}\). The choice \(\pi\ln 3\) incorrectly applies the antiderivative rule for \(\frac{1}{x}\) rather than \(\frac{1}{x^2}\). Always square the radius function before integrating, and be careful to use the correct power rule for the resulting expression.

Net displacement is \(\int_0^3 (t^2-4)\,dt = \left[\frac{t^3}{3}-4t\right]_0^3 = 9-12=-3\), indicating the particle ends up 3 units in the negative direction from its start. The choice \(9\) is wrong because it only accounts for the \(\frac{t^3}{3}\) term and omits the negative contribution of \(-4t\). Net displacement can be negative even when part of the motion is in the positive direction, since it reflects overall position change.

Using integration by parts, \(\int_1^e \ln x\,dx = [x\ln x - x]_1^e = (e-e)-(0-1)=1\). The choice \(e-1\) mistakenly treats the antiderivative as just \(x\ln x\) without subtracting the \(-x\) term correctly at both bounds. Remember that integrating \(\ln x\) requires integration by parts, giving the antiderivative \(x\ln x - x\).

The curves intersect where \(x^3-4x=0\), giving \(x=-2,0,2\); since the cubic is above the line on \((-2,0)\) and below it on \((0,2)\), computing each piece separately and summing the absolute values gives a total area of \(4+4=8\). The choice \(0\) is wrong because it would result from integrating \(x^3-4x\) across the full interval without splitting at the sign change, causing the two regions to cancel. Whenever curves cross more than once, split the integral at each intersection point and sum the absolute areas rather than integrating straight through.

Because the axis of rotation is \(y=-1\), both radii must be shifted upward by 1: the outer radius becomes \((x+2)-(-1)=x+3\) and the inner radius becomes \(x^2-(-1)=x^2+1\), giving \(\pi\int_{-1}^2[(x+3)^2-(x^2+1)^2]\,dx\). The option using \((x+2)^2-(x^2)^2\) is wrong because it forgets to shift the radii to account for the axis being at \(y=-1\) instead of the x-axis. Whenever the axis of rotation is not the x-axis or y-axis, always translate each radius by the appropriate constant before squaring.

The semicircle's radius is half the diameter, \(\frac{\sqrt{x}}{2}\), so its area is \(\frac{1}{2}\pi\left(\frac{\sqrt{x}}{2}\right)^2 = \frac{\pi x}{8}\), and integrating gives \(V=\int_0^4 \frac{\pi x}{8}\,dx = \frac{\pi}{8}\cdot 8 = \pi\). The choice \(4\pi\) mistakenly uses the full diameter as the radius rather than halving it first. When cross-sections are semicircles, always divide the given diameter by two before computing the area formula \(\frac{1}{2}\pi r^2\).

The curves intersect at \(x=1\), and the shell height is \((2-x^2)-x^2=2-2x^2\), so \(V=2\pi\int_0^1 x(2-2x^2)\,dx = 2\pi\left[x^2-\frac{x^4}{2}\right]_0^1 = 2\pi(0.5)=\pi\). The choice \(2\pi\) results from forgetting the factor of \(\frac{1}{2}\) that appears when integrating \(2x^3\). Always simplify the shell height fully before multiplying by the shell radius and the outer factor of \(2\pi\).

The net rate of change is \(R(t)-D(t)=4-t\), which is positive for \(t<4\) and negative for \(t>4\), so the amount of water increases until \(t=4\) and decreases afterward, making \(t=4\) the maximum. The choice \(t=6\) is wrong because at that time the net rate \(4-6=-2\) is already negative, meaning water is already decreasing. To find extrema of an accumulated quantity, set the net rate function equal to zero and check the sign change, just as with any critical point analysis.

Since \(e^x \ge x+1\) for all \(x\) (they are tangent at \(x=0\)), the area is \(\int_0^1 [e^x-(x+1)]\,dx = \left[e^x-\frac{x^2}{2}-x\right]_0^1 = (e-0.5-1)-1 = e-2.5\). The choice \(e-1.5\) results from forgetting to subtract the constant term \(-x\) evaluated at \(x=1\). When the curves involve both exponential and polynomial terms, carefully evaluate each piece of the antiderivative at both bounds before combining.

Rewriting the boundary as \(x=y^2\) for \(0 \le y \le 2\), the volume is \(\pi\int_0^2 (y^2)^2\,dy = \pi\int_0^2 y^4\,dy = \pi\left[\frac{y^5}{5}\right]_0^2=\frac{32\pi}{5}\). The choice \(\frac{32\pi}{3}\) incorrectly uses a power of 3 instead of 5 when integrating \(y^4\). When rotating about the y-axis using disks, remember to rewrite the boundary function in terms of \(y\) before setting up the integral.

The graph of \(y=\sqrt{4-t^2}\) from \(t=0\) to \(t=2\) is a quarter circle of radius 2, whose area is \(\frac{1}{4}\pi(2)^2=\pi\), matching \(F(2)\). The choice \(4\pi\) represents the area of the full circle rather than just the quarter that lies over \([0,2]\). Recognizing standard geometric shapes hidden inside an accumulation function can save significant computation time on the exam.

Since \(v(t)\) is positive on \((0,\pi)\) and negative on \((\pi,2\pi)\), the distance is \(\int_0^\pi 3\sin t\,dt + \left|\int_\pi^{2\pi} 3\sin t\,dt\right| = 6+6=12\). The choice \(0\) is incorrect because that corresponds to the net displacement, where the symmetric positive and negative regions cancel exactly. Always split the interval at each zero of the velocity function and sum the absolute values of each piece to find total distance.

With shell radius \(3-x\) and height \(4-x^2\) over \(x \in [-2,2]\), expanding the product and noting that odd-power terms vanish on the symmetric interval leaves \(V=2\pi\int_{-2}^2(12-3x^2)\,dx = 2\pi(32)=64\pi\). The choice \(32\pi\) omits the leading factor of 2 that comes from the shell method's \(2\pi\) constant. When the axis of rotation is shifted away from the y-axis, remember the shell radius becomes (axis position minus \(x\)), not just \(x\) itself.

Using symmetry over \([-2,2]\), the volume is \(2\pi\int_0^2(16-x^4)\,dx = 2\pi\left(32-\frac{32}{5}\right)=2\pi\cdot\frac{128}{5}=\frac{256\pi}{5}\). The choice \(\frac{128\pi}{5}\) omits the factor of 2 that arises from exploiting the symmetry of the region across the y-axis. When a region is symmetric about the y-axis, doubling the integral from 0 to the positive bound is a useful shortcut, but the factor of 2 must not be dropped.

Since \(g'(x)=f(x)\) by the Fundamental Theorem, \(g\) decreases while \(f\) is negative on \((0,2)\) and increases while \(f\) is positive on \((2,5)\), so the local minimum occurs at \(x=2\); evaluating \(g(2)=\int_0^2 f(t)\,dt=-3\) gives the minimum value. The choice \(g(2)=7\) incorrectly uses the value of the second given integral rather than the accumulated value up to \(x=2\). To locate extrema of an accumulation function, treat \(f(x)\) as the derivative and apply the first derivative test just as with any other function.

Since \(2 \ge \sec^2 x\) on \([0,\pi/4]\), the area is \(\int_0^{\pi/4}(2-\sec^2x)\,dx = [2x-\tan x]_0^{\pi/4} = \left(\frac{\pi}{2}-1\right)-0=\frac{\pi}{2}-1\). The choice \(\frac{\pi}{4}-1\) incorrectly uses \(\frac{\pi}{4}\) instead of doubling the bound after integrating the constant term \(2\). Remember that \(\int \sec^2 x\,dx = \tan x\), a standard antiderivative that frequently appears in area and volume problems involving trigonometric functions.