Applications of Integration

The average value 平均值 of a continuous function 连续函数 $f$ over $[a,b]$ is the integral 积分 divided by the interval 区间 length:

It is the constant height of a rectangle 矩形 on $[a,b]$ with the same area as under $f$. Do not confuse it with the average rate of change 平均变化率 (which divides change in $f$ by the interval), or with a Riemann sum 黎曼和 average of a few values – the exam distinguishes these carefully. Report units in context (e.g. the average rate “in vehicles per hour”).

Worked example. The average value of $f(x)=x^2$ on $[0,3]$ is

Integration reverses the motion links of Unit 4. For a particle 质点 in straight-line motion over $[t_1,t_2]$:

• Displacement 位移 (net change in position) $= \displaystyle\int_{t_1}^{t_2} v(t)\,dt$.
• Total distance travelled 总路程 $= \displaystyle\int_{t_1}^{t_2} |v(t)|\,dt$ – integrate speed 速率, so direction changes add up instead of cancelling.
• Position at a later time $= x(t_1) + \displaystyle\int_{t_1}^{t} v(s)\,ds$; velocity 速度 from acceleration 加速度 $v(t)=v(t_1)+\int_{t_1}^t a(u)\,du$.

The displacement-versus-distance distinction (with the absolute value 绝对值) is a classic graded point.

Read the sign before adding motion

For $v(t)=t-2$ m/s on $[0,3]$, solve $v(t)=0$ and test either side. The particle moves left on $[0,2)$ and right on $(2,3]$.

Displacement is signed:

Total distance counts both journeys positively. Split at the sign change 符号变化: The particle travels two metres left and half a metre right. Taking only $|\int_0^3v(t)\,dt|=1.5$ m loses the return journey.

Position needs a starting position. If $x(0)=5$ m, then

The final position is right of the origin 原点, even though displacement is negative.

A zero of velocity is a candidate for a turn. Check signs: $v(t)=(t-1)^2$ is positive on both sides of $t=1$, so the particle stops momentarily without changing direction.

Recover velocity with an initial condition

For $a(t)=4$ m/s$^2$ and $v(0)=1$ m/s:

Here $t$ is in seconds and $v(t)$ is in metres per second. Equivalently, integrate to $4t+C$ and use $v(0)=1$ to find $C=1$. Each integration needs its own initial condition 初始条件.

Original exam applications

2025 AB Q5D (calculator not allowed). The original gives $v_J(t)=2t(t^2-1)^3$ and $x_J(0)=7$, without physical units. Recover position, using $u=t^2-1$ and $du=2t\,dt$:

The change in position is $20$, while final position is $27$. Keep the initial position outside the integral.

2014 AB Q4(c). At times $2,5,8,12$ min, train velocities are $100,40,-120,-150$ m/min. Starting position is $s_A(2)=300$ m east of the origin. Write the exact position expression, then estimate the signed change by the three specified trapezoids 梯形:

This means about $150$ m west of the origin. Preserve the unequal interval widths and signs. Sampled velocities do not locate every zero, so absolute trapezoids do not give exact total distance.

A function defined as an integral accumulates a rate of change. The net change 净变化 theorem is the everyday tool: the definite integral 定积分 of a rate over an interval gives the net change of the quantity:

So “how much water is in the tank at $t=5$” = starting amount + $\int_0^5(\text{inflow}-\text{outflow})\,dt$. Watch signs (in minus out) and units. Many multi-part FRQs are built entirely on this idea – often paired with “write, but do not evaluate, an integral expression that gives the total...”.

Worked example (rate in, rate out). Water flows into a pipe at $12$ cubic feet per hour and drains out at $D(t)=3t$ cubic feet per hour, for $0\le t\le8$. At $t=0$ the pipe holds $30$ cubic feet.

(a) Amount. $W(t)=30+\displaystyle\int_0^t(12-3s)\,ds=30+12t-1.5t^2$.

(b) Increasing or decreasing at $t=3$? Compare the rates, not the amounts: $W'(3)=12-3(3)=3>0$, so the amount is increasing, because water enters faster than it leaves.

(c) Minimum amount. $W'(t)=0$ at $t=4$. Use the candidates test – check the critical point and both endpoints:

$t$	$0$	$4$	$8$
$W(t)$	$30$	$54$	$30$

The minimum is $30$ cubic feet, at $t=0$ and $t=8$; $t=4$ is the maximum. A sign chart of $W'$ alone can find the maximum, but only the table settles which endpoint is lowest.

The area between two curves $y=f(x)$ (top) and $y=g(x)$ (bottom) on $[a,b]$ is:

Find the intersection points 交点 first to set the limits of integration 积分限, and always subtract bottom from top.

Worked example. Find the area between $y=x$ and $y=x^2$. They cross at $x=0$ and $x=1$, and on $(0,1)$ the line $y=x$ is on top, so

Some regions are easier described with horizontal slices 水平切片 – integrate with respect to $y$:

Choose $x$- or $y$-slices to keep a single top/bottom (or right/left) pair over the whole region.

Worked example. Find the same area between $y=x$ and $y=x^2$ with horizontal slices. Solve each curve for $x$: the line is $x=y$ and the parabola 抛物线 is $x=\sqrt{y}$. For $0<y<1$, $\sqrt{y}>y$ (try $y=0.25$: $0.5>0.25$), so the parabola is on the right:

Both methods give $\tfrac16$, as they must.

If the curves cross more than twice, one function is on top for part of the region and the other on top elsewhere. Split into a sum of integrals at each crossing, or integrate the absolute value of the difference:

Determine which curve is higher on each subinterval before writing the integrals.

Worked example. Find the area between $y=x^3$ and $y=x$. They meet where $x^3=x$, at $x=-1,0,1$. Test a point in each piece: at $x=-\tfrac12$, $x^3=-\tfrac18>-\tfrac12$, so the cubic is on top; at $x=\tfrac12$ the line is on top.

Common mistake. Writing $\displaystyle\int_{-1}^{1}(x-x^3)\,dx=0$: the two pieces cancel. On the calculator, $\displaystyle\int_{-1}^{1}\left|x-x^3\right|dx$ gives $\tfrac12$ in one step.

If a solid has a known cross section 横截面 perpendicular 垂直 to an axis, its volume 体积 is the integral of the cross-sectional area:

For square 正方形 cross sections with side equal to the region's height $s(x)=f(x)-g(x)$, use $A(x)=[s(x)]^2$; for a rectangle of height $k\cdot s(x)$, use $A(x)=k\,[s(x)]^2$. The exam phrase “the base of a solid... cross sections are squares/rectangles” signals exactly this.

Worked example. The base of a solid is the region under $y=\sqrt{x}$ from $x=0$ to $x=4$, and cross sections perpendicular to the $x$-axis are squares (the solid in the figure below). Each square's side is $s(x)=\sqrt{x}$, so $A(x)=x$ and

With semicircles on the same base instead, $A(x)=\tfrac{\pi}{8}\big(\sqrt{x}\big)^2$, so $V=\tfrac{\pi}{8}\cdot 8=\pi$: only the area formula changes.

Worked example. The base of a solid is the region between $y=x$ and $y=x^2$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

The side of each square is the slice length $s(x)=x-x^2$, so $A(x)=(x-x^2)^2=x^2-2x^3+x^4$ and

Expand the square before integrating; $\int (x-x^2)^2\,dx$ is not $\left(\int (x-x^2)\,dx\right)^2$.

Same method, different area formula: for an equilateral triangle 三角形 of side $s$, $A=\tfrac{\sqrt{3}}{4}s^2$; for a semicircle 半圆 of diameter 直径 $s$, $A=\tfrac{\pi}{8}s^2$. Set the shape's key length equal to the region's slice length $s(x)$, then integrate $A(x)$.

Worked example. The base is the region between $y=x$ and $y=x^2$ for $0\le x\le1$. Cross sections perpendicular to the $x$-axis have side (or diameter) $s=x-x^2$, and the squares section found $\displaystyle\int_0^1(x-x^2)^2\,dx=\frac1{30}$.

• Equilateral triangles: $V=\dfrac{\sqrt3}{4}\displaystyle\int_0^1(x-x^2)^2\,dx=\dfrac{\sqrt3}{4}\cdot\dfrac1{30}=\dfrac{\sqrt3}{120}$.
• Semicircles: $V=\dfrac{\pi}{8}\cdot\dfrac1{30}=\dfrac{\pi}{240}$.

Only the constant in front changes: square $1$, equilateral triangle $\tfrac{\sqrt3}{4}$, semicircle $\tfrac{\pi}{8}$. The semicircle area is $\tfrac12\pi r^2$ with radius $r=\tfrac s2$: $\tfrac12\pi\tfrac{s^2}{4}=\tfrac{\pi}{8}s^2$. Using $s$ as the radius is the usual slip.

Revolving a region around an axis makes a solid of revolution 旋转体. If the region touches the axis, each slice is a disc 圆盘 of radius 半径 $r$ = the function value:

Around the $x$-axis, $r=f(x)$ and integrate in $x$; around the $y$-axis, express $x$ as a function of $y$ and integrate in $y$.

Worked example. Revolve the region under $y=\sqrt{x}$ from $x=0$ to $x=4$ about the $x$-axis. Each disc has radius $r=\sqrt{x}$, so

Around any horizontal or vertical line $y=k$ or $x=k$, the radius is the distance from the curve to that line, e.g. $r(x)=|f(x)-k|$. Set up the radius carefully, then use $V=\pi\int r^2$.

Worked example. The region between $y=\sqrt x$ and the line $y=2$, for $0\le x\le4$, is revolved about $y=2$. Each disc has radius $R=2-\sqrt x$ (line minus curve, since the line is above):

For a write, but do not evaluate question, the integral with correct limits and radius is the whole answer – do not waste time evaluating it.

When the region does not touch the axis, each slice is a washer 垫圈 (a ring) with an outer radius $R$ and inner radius $r$:

$R$ is the distance from the axis to the farther boundary and $r$ to the nearer one. Around a line other than an axis, both radii are measured as distances to that line – a very common exam variation (e.g. revolving about $y=-2$).

Worked example. Revolve the region between $y=\sqrt{x}$ and $y=x^2$ ($0\le x\le 1$) (the region in the figure) about (a) the $x$-axis and (b) the line $y=-1$.

(a) On $(0,1)$, $\sqrt{x}>x^2$, so $y=\sqrt{x}$ is farther from the $x$-axis: $R=\sqrt{x}$ and $r=x^2$.

(b) Each radius is now a distance to $y=-1$, so add $1$ to each height: $R=\sqrt{x}+1$ and $r=x^2+1$. Expand both squares:

Common mistakes. Writing $\pi\int (R-r)^2\,dx$ instead of $\pi\int (R^2-r^2)\,dx$; and, about $y=-1$, using $R=\sqrt{x}-1$ (subtracting) instead of $R=\sqrt{x}-(-1)=\sqrt{x}+1$.

• Area between curves is $\int(\text{top}-\text{bottom})\,dx$ — find the intersection points for the limits and keep top minus bottom.
• For a volume of revolution, add up disc/washer cross-sections of area $\pi r^2$ (or $\pi(R^2-r^2)$).
• The average value of $f$ on $[a,b]$ is $\tfrac{1}{b-a}\int_a^b f\,dx$.
• Accumulated change is $\int$ of a rate: total = initial value $+\int_a^b(\text{rate})\,dt$.
• Integration means "adding up infinitely many tiny pieces" — set up the integrand as one thin slice.

Practice.

1. Find the average value of $f(x)=\sin x$ on $[0,\pi]$.
2. $v(t)=2t-4$ m/s for $0\le t\le5$. Find the displacement and the total distance.
3. Find the area enclosed by $y=x^2$ and $y=4$.
4. Find the volume when the region under $y=x^2$, $0\le x\le2$, is revolved about the $x$-axis.
5. The region between $y=x$ and $y=x^2$ is revolved about the line $y=-1$. Write, but do not evaluate, an integral for the volume.

Answers.

1. $\dfrac1\pi\displaystyle\int_0^\pi\sin x\,dx=\dfrac{2}{\pi}$.
2. Displacement $\displaystyle\int_0^5(2t-4)\,dt=25-20=5$ m. $v<0$ on $(0,2)$: distance $=4+9=13$ m.
3. They meet at $x=\pm2$: $\displaystyle\int_{-2}^2(4-x^2)\,dx=16-\frac{16}{3}=\frac{32}{3}$.
4. $\pi\displaystyle\int_0^2x^4\,dx=\dfrac{32\pi}{5}$.
5. $\pi\displaystyle\int_0^1\left[(x+1)^2-(x^2+1)^2\right]dx$.