Applications of Integration

If $f(x)\ge g(x)$ on $[a,b]$, the area of the region between the curves is

If instead the region is bounded left/right by $x=v(y)$ (right) and $x=u(y)$ (left) for $c\le y\le d$, integrate in $y$:

The average value of $f$ on $[a,b]$ is

By the Mean Value Theorem for Integrals, if $f$ is continuous there is at least one $c\in(a,b)$ with $f(c)=f_{\text{avg}}$.

If $F'(t)=r(t)$ is a rate of change, the net change of $F$ over $[a,b]$ is

For a particle with velocity $v(t)$:

Revolving a region about a horizontal axis and slicing perpendicular to it gives circular cross-sections.

where $R$ is the outer radius (far edge from axis) and $r$ the inner radius (near edge). For revolution about a vertical axis, integrate in $y$ with radii measured horizontally.

Slicing parallel to the axis of revolution produces thin cylindrical shells:

For revolution about the $y$-axis with a region described by $x$: radius $=x$, height $=$ top $-$ bottom. For revolution about the $x$-axis, integrate in $y$: radius $=y$, height $=$ right $-$ left.

If every slice perpendicular to the $x$-axis has area $A(x)$, then

Common cross-sections built on a segment of length $s$ (usually $s=\text{top}-\text{bottom}$ of the base region):

The length of the smooth curve $y=f(x)$ from $x=a$ to $x=b$ is

It comes from summing hypotenuses $\sqrt{(dx)^2+(dy)^2}=\sqrt{1+(dy/dx)^2}\,dx$ of tiny right triangles along the curve.

Revolving $y=f(x)\ge 0$ on $[a,b]$ about the $x$-axis produces a surface of area

About the $y$-axis (region in the first quadrant), $S=\displaystyle\int 2\pi\,x\,\sqrt{1+(y')^2}\,dx$. Think “circumference $2\pi r$ times arc-length element.”

Work done by a variable force $F(x)$ moving an object from $x=a$ to $x=b$ is

For a spring, Hooke's law gives $F(x)=kx$, so $W=\int_0^d kx\,dx=\tfrac12 kd^2$.

To pump fluid out of a tank, slice the fluid into horizontal layers at height $y$ of area $A(y)$ and thickness $dy$. A layer has weight $\rho g\,A(y)\,dy$ and must be lifted a distance $D(y)$:

where $\rho g$ is the fluid's weight density (for water, $\rho g\approx 9800\ \text{N/m}^3$, or $62.4\ \text{lb/ft}^3$).

Match the slice to the axis:

• Slice perpendicular to the axis $\Rightarrow$ disks/washers; the variable of integration matches the axis direction.
• Slice parallel to the axis $\Rightarrow$ shells.

Prefer whichever keeps your radii/heights as single functions and avoids solving the boundary curve for the other variable. Revolving a region under $y=f(x)$ about the $y$-axis is usually one clean shell integral $2\pi\int x f(x)\,dx$, whereas washers would require inverting $f$.

For a curve given by $x=x(t)$, $y=y(t)$ on $[\alpha,\beta]$, the arc length is

Example: for $x=\cos t$, $y=\sin t$, $\;\sqrt{(-\sin t)^2+(\cos t)^2}=1$, so one full loop $t\in[0,2\pi]$ has length $\int_0^{2\pi}1\,dt=2\pi$ — the circumference of the unit circle, as expected. This unifies with $y=f(x)$: taking $t=x$ recovers $\sqrt{1+(dy/dx)^2}$.