积分的应用

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

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

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

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.

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

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:

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.

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

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.

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 $\pi(\text{radius})^2$:

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.

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:

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

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.