Applications of Integration

The average value 平均值 of $f$ over $[a,b]$ is the integral 积分 divided by the width:

It is the constant height a rectangle 矩形 would need to have the same area as the region under $f$. Do not confuse this with the average rate of change 变化率 (which uses the derivative 导数).

Worked example. The average value of $f(x)=x^2$ on $[0,3]$ is $\dfrac{1}{3}\displaystyle\int_0^3 x^2\,dx=\dfrac13\left[\dfrac{x^3}{3}\right]_0^3=\dfrac13(9)=3$.

Worked example (average from a table). The height $H$ of a tree, in metres, is measured at times $t$ in years.

$t$	$2$	$3$	$5$	$7$	$10$
$H(t)$	$1.5$	$2$	$6$	$11$	$15$

Use a trapezoidal sum with the four subintervals to estimate the average height for $2\le t\le10$. The widths are unequal ($1,2,2,3$), so compute each trapezoid separately:

The average height is about $\dfrac{65.75}{10-2}\approx8.22$ metres. Forgetting to divide by $b-a=8$ is the most common error.

When is the function equal to its average? If a question asks for the time $t$ at which $A(t)$ equals its average value on $[0,30]$, first compute $A_{\text{avg}}=\dfrac1{30}\displaystyle\int_0^{30}A(t)\,dt$ on the calculator, then solve $A(t)=A_{\text{avg}}$ with the calculator. The Mean Value Theorem for integrals promises such a $t$ exists when $A$ is continuous.

For straight-line motion, integration 积分法 reverses differentiation 求导:

Two key distinctions: displacement 位移 is $\int_a^b v\,dt$ (net change 净变化 in position), while total distance 总路程 is $\int_a^b |v|\,dt$ (splitting where $v$ changes sign). Speed 速率 is $|v|$.

Worked example. A particle has $v(t)=t^2-4t+3$ for $0\le t\le3$ (metres per second).

• Displacement: $\displaystyle\int_0^3(t^2-4t+3)\,dt=\left[\frac{t^3}{3}-2t^2+3t\right]_0^3=9-18+9=0$ m. The particle ends where it started.
• Distance: $v=(t-1)(t-3)$ is negative on $(1,3)$. $\displaystyle\int_0^1v\,dt=\frac43$ and $\displaystyle\int_1^3v\,dt=-\frac43$, so the total distance is $\dfrac43+\dfrac43=\dfrac83$ m.

Position from a known point. If a train is $300$ m east of the station at $t=2$, its position is $x(t)=300+\displaystyle\int_2^t v(s)\,ds$. The lower limit is the time you know the position, not always $0$.

Interpreting with units. "$\displaystyle\int_0^{40}|v(t)|\,dt$" with $v$ in metres per second means "the total distance, in metres, that the object travels from $t=0$ to $t=40$ seconds". Name the quantity, the units and the time interval.

When a rate is given (flow rate, sales per day), the definite integral 定积分 gives the accumulated total 累积总量, and $\int_a^b R(t)\,dt$ carries the units of $R$ times time. A common setup: initial amount 初始量 $+\int(\text{rate in}-\text{rate out})\,dt$ gives the amount at a later time. Always interpret the answer in context, with units.

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 $y=f(x)$ (top) and $y=g(x)$ (bottom) from $a$ to $b$ is

Find the intersection points 交点 for the limits of integration 积分限, and always subtract top minus bottom.

Worked example. Between $y=x$ and $y=x^2$ (crossing at $x=0,1$, with $y=x$ on top), the area is $\displaystyle\int_0^1 (x-x^2)\,dx=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac16$.

When curves are easier to describe as $x=f(y)$, integrate with respect to $y$ instead, using right minus left:

Choosing to integrate in $y$ can avoid splitting the region into several pieces.

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 two curves cross several times, the top and bottom switch. Split the region at each intersection and integrate each piece with the correct top-minus-bottom (or use $\int|f-g|$), then add the pieces.

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's cross sections 横截面 perpendicular 垂直 to the $x$-axis are squares 正方形 or rectangles, integrate their area to find the volume 体积. With side length equal to the distance between two curves, a square cross section gives

The method is always "integrate the cross-sectional area."

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. 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 idea, different area formula: for equilateral triangle 等边三角形 cross sections use $A=\tfrac{\sqrt3}{4}s^2$, and for semicircular 半圆形的 ones $A=\tfrac{\pi}{8}s^2$ (with $s$ the distance between the curves). Substitute the area formula and integrate.

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 whose cross sections are discs. The disc method 圆盘法 integrates $\pi(\text{radius})^2$:

where the radius 半径 $R$ is the distance from the curve to the axis. Use $dy$ when revolving around the $y$-axis.

Worked example. Revolving the region under $y=\sqrt{x}$ from $0$ to $4$ about the $x$-axis gives discs of radius $\sqrt{x}$: $V=\pi\displaystyle\int_0^4 (\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=8\pi$.

When the axis of revolution 旋转轴 is a horizontal or vertical line like $y=k$ (not an axis), the radius adjusts: $R=|f(x)-k|$. Set up the radius as the distance from the curve to that line, then integrate $\pi R^2$ as before.

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.

If the region does not touch the axis, revolving leaves a hole, so cross sections are washers (rings 圆环). The washer method 垫圈法 subtracts the inner disc:

Identify the outer and inner radii as distances from each curve to the axis.

As with discs, revolving around a line $y=k$ or $x=k$ shifts both radii – each becomes the distance from its curve to that line. Sketch the region and the axis, label $R_{\text{outer}}$ and $R_{\text{inner}}$, then integrate the difference of squares.

Worked example. Revolve the region between $y=\sqrt{x}$ and $y=x^2$ ($0\le x\le 1$) 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 $R=\sqrt{x}-(-1)=\sqrt{x}+1$.

The arc length 弧长 of $y=f(x)$ from $a$ to $b$ is

This BC-only formula comes from summing tiny hypotenuses 斜边 $\sqrt{dx^2+dy^2}$. The same idea gives the distance a particle 质点 travels along a curved path.

Worked example (perimeter). $R$ is the region between $y=x$ and $y=x^2$ for $0\le x\le1$. Its boundary is a straight segment from $(0,0)$ to $(1,1)$, of length $\sqrt2$, and the parabola, of length $\displaystyle\int_0^1\sqrt{1+(2x)^2}\,dx$. So

Arc-length integrals rarely have a neat antiderivative; set them up and use the calculator.

Worked example. Find the arc length of $y=\tfrac{2}{3}x^{3/2}$ from $x=0$ to $x=3$. Here $f'(x)=x^{1/2}$, so $1+(f')^2=1+x$ and

• 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.
• Total distance uses $|v|$; displacement uses $v$. For a minimum or maximum amount, use the candidates test with the endpoints.

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.
6. Write an integral for the length of $y=\ln x$ from $x=1$ to $x=e$.

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. $\pi\displaystyle\int_0^1\left[(x+1)^2-(x^2+1)^2\right]dx$.
6. $\displaystyle\int_1^e\sqrt{1+\frac1{x^2}}\,dx$.