Applications of Integration

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

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

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 reverses differentiation:

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

When a rate is given (flow rate, sales per day), the definite integral gives the accumulated total, and $\int_a^b R(t)\,dt$ carries the units of $R$ times time. A common setup: initial amount $+\int(\text{rate in}-\text{rate out})\,dt$ 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

Find the intersection points for the limits, and always subtract top minus bottom.

The area between two curves is the integral of 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.

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 to the $x$-axis are squares or rectangles, integrate their area. With side length equal to the distance between two curves, a square cross section gives

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

Same idea, different area formula: for equilateral-triangle cross sections use $A=\tfrac{\sqrt3}{4}s^2$, and for semicircular 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 whose cross sections are discs. The disc method 圆盘法 integrates $\pi(\text{radius})^2$:

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

The disc method: rotating y=f(x) about the axis sweeps out disks of radius f(x)

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

Rotating a region about an axis sweeps out a solid of revolution

When the axis of revolution 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). 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.

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

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

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

The bridge's main cable hangs as a smooth curve; an integral gives its exact length

• 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.