Applications of Integration
Unit 8 of AP Calculus AB, worth 10–15% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Average value of a function, motion problems, accumulation in context, area between curves, volumes by cross-sections and by discs and washers.
How this unit is tested
Every problem in this unit reduces to the same question: what does a definite integral of this particular function actually accumulate? Before setting up any integral, identify the units of the integrand and the units of the answer — that tells you whether you're finding an average, a net change, an area, or a volume. For motion, always distinguish velocity (rate of position change) from speed (|velocity|); displacement uses the signed integral of v(t), total distance requires you to split the interval at the zeros of v(t) and integrate |v(t)| on each piece.
For area and volume problems, start by sketching or at least describing the region and finding where the bounding curves intersect — these intersection points become your limits of integration. Decide whether vertical strips (integrate dx, use top minus bottom) or horizontal strips (integrate dy, use right minus left) make the algebra easier; a region that requires two integrals in x may need only one in y.
For volumes, identify the cross-sectional shape perpendicular to the axis of revolution or slicing axis at a generic value of x (or y). If the cross-section is a disc, $V=\pi\int R(x)^2\,dx$; if it has a hole, it's a washer, $V=\pi\int [R(x)^2-r(x)^2]\,dx$ — never $\pi\int (R(x)-r(x))^2\,dx$, that's a different (wrong) shape. If the solid is rotated about a line other than the x- or y-axis, adjust R and r by subtracting or adding the location of that line before squaring.
Finally, for average value and the Mean Value Theorem for Integrals, remember the average value is not $\int_a^b f(x)\,dx$ — you must divide by $(b-a)$. MVT then guarantees some c in the interval where f actually equals that average, which is a separate step (setting $f(c)$ equal to the average and solving) from computing the average itself.
What you have to know
Average value of a function
The average value of f on $[a,b]$ is $f_{avg}=\dfrac{1}{b-a}\int_a^b f(x)\,dx$.
Mean Value Theorem for Integrals
If f is continuous on $[a,b]$, there exists at least one c in $(a,b)$ such that $f(c)=f_{avg}=\dfrac{1}{b-a}\int_a^b f(x)\,dx$.
Displacement vs. total distance
For velocity $v(t)$ on $[a,b]$: displacement $=\int_a^b v(t)\,dt$; total distance $=\int_a^b |v(t)|\,dt$, computed by splitting at zeros of $v(t)$ and summing the absolute values of the signed integrals on each subinterval.
Area between two curves
If $f(x)\ge g(x)$ on $[a,b]$, the area between them is $\int_a^b [f(x)-g(x)]\,dx$; analogously for horizontal strips, $\int_c^d [f(y)-g(y)]\,dy$.
Volume by cross-sections
If $A(x)$ is the cross-sectional area of a solid perpendicular to the x-axis, its volume is $V=\int_a^b A(x)\,dx$.
Volume by discs and washers
Rotating about an axis: disc method $V=\pi\int_a^b R(x)^2\,dx$; washer method (with an inner radius $r(x)$) $V=\pi\int_a^b [R(x)^2-r(x)^2]\,dx$.
14 practice questions
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Find the average value of $f(x)=x^2$ on $[0,3]$.
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Answer. 3
Average value is $\frac{1}{3-0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3=\frac{1}{3}(9)=3$. Divide by the length of the interval, not just compute the integral. -
Find the average value of $f(x)=\sin x$ on $[0,\pi]$.
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Answer. $\dfrac{2}{\pi}$
$f_{avg}=\frac{1}{\pi}\int_0^\pi \sin x\,dx=\frac{1}{\pi}[-\cos x]_0^\pi=\frac{1}{\pi}(1-(-1))=\frac{2}{\pi}$. -
A particle moves with velocity $v(t)=t^2-4t+3$ (in meters/sec) for $0\le t\le 4$. Find both the displacement and the total distance traveled.
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Answer. Displacement = 4/3 m; total distance = 4 m
v(t)=(t-1)(t-3) changes sign at t=1,3. On [0,1] and [3,4] v>0 with integral 4/3 each; on [1,3] v<0 with integral −4/3. Displacement sums these signed values to 4/3; total distance sums their absolute values to 4/3+4/3+4/3=4. -
A particle has velocity $v(t)=6t-t^2$ and initial position $s(0)=2$. Find its position at $t=5$.
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Answer. $s(5)=\dfrac{106}{3}$
$s(5)=s(0)+\int_0^5(6t-t^2)\,dt = 2+\left[3t^2-\frac{t^3}{3}\right]_0^5=2+(75-\frac{125}{3})=2+\frac{100}{3}=\frac{106}{3}$. -
For a particle with position governed by velocity $v(t)=3t^2-12t+9$, determine whether the particle is speeding up or slowing down at $t=0.5$.
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Answer. Slowing down
$v(0.5)=3(0.25)-6+9=3.75>0$. Acceleration $a(t)=6t-12$, so $a(0.5)=3-12=-9<0$. Since velocity and acceleration have opposite signs, the particle is slowing down. -
Water flows into a tank at rate $r(t)=10-0.5t$ gal/min and is pumped out at a constant 4 gal/min, for $0\le t\le20$. The tank starts with 50 gallons. Find the amount of water in the tank at $t=10$, and state whether it is increasing or decreasing at that moment.
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Answer. 85 gallons; increasing
Net rate is $r(t)-4=6-0.5t$. Amount $=50+\int_0^{10}(6-0.5t)\,dt=50+[6t-0.25t^2]_0^{10}=50+35=85$. Net rate at $t=10$ is $6-5=1>0$, so the amount is increasing. -
People enter a stadium at rate $E(t)=200t-10t^2$ people per minute for $0\le t\le20$. What does $\int_0^{10}E(t)\,dt$ represent, and what is its value?
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Answer. It represents the total number of people who entered in the first 10 minutes; value = 20000/3 ≈ 6666.7 people
$\int_0^{10}(200t-10t^2)\,dt=\left[100t^2-\frac{10}{3}t^3\right]_0^{10}=10000-\frac{10000}{3}=\frac{20000}{3}$. Integrating a rate over time gives the accumulated total quantity, here total people entered. -
Find the c guaranteed by the Mean Value Theorem for Integrals for $f(t)=t^3-2t$ on $[0,2]$.
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Answer. $c=\sqrt{2}$
Average value is $\frac{1}{2}\int_0^2(t^3-2t)\,dt=\frac{1}{2}\left[\frac{t^4}{4}-t^2\right]_0^2=\frac{1}{2}(4-4)=0$. Solve $f(c)=0$: $c^3-2c=c(c^2-2)=0$, giving $c=0,\pm\sqrt2$; only $c=\sqrt2$ lies in $(0,2)$. -
Find the area of the region enclosed by $y=x^2$ and $y=-x^2+4x$.
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Answer. 8/3
Set equal: $x^2=-x^2+4x \Rightarrow 2x^2-4x=0 \Rightarrow x=0,2$. Area $=\int_0^2[(-x^2+4x)-x^2]\,dx=\int_0^2(4x-2x^2)\,dx=[2x^2-\frac{2}{3}x^3]_0^2=8-\frac{16}{3}=\frac{8}{3}$. -
Find the area of the region enclosed by $x=y^2$ and $x=y+2$.
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Answer. 9/2
Set $y^2=y+2\Rightarrow (y-2)(y+1)=0$, so $y=-1,2$. Since $x=y+2$ is to the right of $x=y^2$ on this interval, area $=\int_{-1}^{2}[(y+2)-y^2]\,dy=\left[\frac{y^2}{2}+2y-\frac{y^3}{3}\right]_{-1}^{2}=\frac{10}{3}-(-\frac{7}{6})=\frac{9}{2}$. Using horizontal strips avoids splitting the region into two x-integrals. -
The region bounded by $y=\sqrt{x}$, $y=0$, and $x=4$ is rotated about the x-axis. Find the volume of the resulting solid.
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Answer. $8\pi$
Disc method: $V=\pi\int_0^4(\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=\pi\left[\frac{x^2}{2}\right]_0^4=8\pi$. -
The region bounded by $x=y^2$, $y=0$, and $y=2$ (with $x\ge0$) is rotated about the y-axis. Find the volume of the resulting solid.
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Answer. $\dfrac{32\pi}{5}$
Since rotation is about the y-axis and x is already expressed in terms of y, use $V=\pi\int_0^2 (y^2)^2\,dy=\pi\int_0^2 y^4\,dy=\pi\left[\frac{y^5}{5}\right]_0^2=\frac{32\pi}{5}$. -
The base of a solid is the region enclosed by $y=x^2$ and $y=4$. Cross-sections perpendicular to the x-axis are squares. Find the volume.
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Answer. $\dfrac{512}{15}$
The curves meet at $x=\pm2$. The side length of each square is the vertical distance $4-x^2$, so $A(x)=(4-x^2)^2$. By symmetry, $V=2\int_0^2(4-x^2)^2\,dx=2\int_0^2(16-8x^2+x^4)\,dx=2\left[16x-\frac{8}{3}x^3+\frac{x^5}{5}\right]_0^2=2\left(32-\frac{64}{3}+\frac{32}{5}\right)=\frac{512}{15}$. -
The region bounded by $y=\sqrt{x}$ and $y=0$ on $[0,4]$ is rotated about the line $y=-1$. Find the volume of the resulting solid.
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Answer. $\dfrac{56\pi}{3}$
Because the axis $y=-1$ lies below the region, this is a washer with outer radius $R(x)=\sqrt{x}+1$ and inner radius $r(x)=1$. $V=\pi\int_0^4[(\sqrt{x}+1)^2-1]\,dx=\pi\int_0^4(x+2\sqrt{x})\,dx=\pi\left[\frac{x^2}{2}+\frac{4}{3}x^{3/2}\right]_0^4=\frac{56\pi}{3}$.
What people get wrong
- Computing $\int_a^b f(x)\,dx$ and calling it the average value — you must also divide by $(b-a)$; write the formula down every time before plugging in.
- Finding total distance by just taking $\left|\int_a^b v(t)\,dt\right|$ — this only gives displacement's magnitude. Instead find every zero of $v(t)$ in $[a,b]$, integrate on each subinterval, and add the absolute values.
- Setting up a washer as $\pi\int (R-r)^2\,dx$ instead of $\pi\int (R^2-r^2)\,dx$ — these are algebraically different. Always square the two radii separately, then subtract.
- Forgetting to shift the radius when rotating about a line other than the x- or y-axis (e.g. using $R(x)=\sqrt{x}$ instead of $\sqrt{x}+1$ when the axis is $y=-1$). Always measure radius as distance from the axis of rotation to the curve.
- In area-between-curves problems, integrating (curve1 − curve2) without checking which function is on top over the whole interval, producing a negative area when curves cross. Split the integral at any intersection point inside $[a,b]$ where the top/bottom relationship switches.
- For speeding up/slowing down questions, checking only the sign of velocity instead of comparing the signs of velocity and acceleration. The particle speeds up when $v(t)$ and $a(t)$ have the same sign, and slows down when they have opposite signs.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Calculus AB. Free, and no account needed to start.