Calculus 2

6 Applications of Integrals

Consider various applications of integration such as computing volume, arc length, surface area, work, hydrostatic force, centroids of planar regions, and applications to business, economics, and life sciences.

6.1 Area Between Two Curves

Area of regions, integrating along the xx-axis

Let f(x)f(x) and g(x)g(x) be continuous functions such that f(x)g(x)f(x) \geq g(x) over an interval [a,b][a, b]. Let R\text{R} denote the region bounded above by f(x)f(x), below by g(x)g(x), on the left by x=ax = a, and on the right by x=bx = b. The area of R\text{R} is given by

A=ab[f(x)g(x)]dxA = \int_{a}^{b} [f(x) - g(x)]dx

Area of compound regions

If we want to look at regions bounded by the graphs of functions that cross one another, we modify the formula by using the absolute value function.

A=abf(x)g(x)dxA = \int_{a}^{b} |f(x) - g(x)|dx

In practice, you must find the points of intersection of the curves and divide the integrals according to the leading and trailing curve.

A=az[f(x)g(x)]dx+zb[f(x)g(x)]dxA = \int_{a}^{z} [f(x) - g(x)]dx + \int_{z}^{b} [f(x) - g(x)]dx

Area of regions, integrating along the yy-axis

Let u(y)u(y) and v(y)v(y) be continuous functions such that u(y)v(y)u(y) \geq v(y) over an interval [c,d][c, d] along the yy-axis. A horizontal strip has a width of u(y)v(y)u(y) - v(y). Thus, the integration formula for area is

A=cd[u(y)v(y)]dyA = \int_{c}^{d} [u(y) - v(y)]dy

6.2 Volume

Area methods can be modified to compute the volume of three-dimensional solids.

The volume of a solid with a known cross-sectional area perpendicular to the xx-axis.

V=abA(x)dxV = \int_{a}^{b} A(x) dx

The disk method is used to find a volume generated when a region is revolved about an axis perpendicular to an approximating strip.

V=abπ[f(x)]2dxV = \int_{a}^{b} \pi [f(x)]^2 dx

The washer method is used to find a volume generated when a region between two curves is revolved about an axis perpendicular to an approximating strip.

V=abπ([f(x)]2[g(x)]2)dxV = \int_{a}^{b} \pi ([f(x)]^2 - [g(x)]^2) dx

The cylindrical shells method is used to find a volume generated when a region is revolved about an axis parallel to an approximating strip.

V=ab2πx f(x)dxV = \int_{a}^{b} 2 \pi x \ f(x) dx

6.3 Polar Forms and Area

The Polar coordinate system

In the polar coordinate system, points are plotted in relation to a fixed point OO, called the origin or pole and a fixed ray emanating from the origin called the polar axis. We associate with each point PP in the plane an ordered pair of numbers P(r,θ)P(r, \theta), where rr (radial coordinate) is the distance from OO to PP and θ\theta (polar angle) is the angle measured from the polar axis.

Changing coordinates

To convert polar coordinates to Cartesian coordinates:

  • x=rcos(θ)x = r\cos(\theta)
  • y=rsin(θ)y = r\sin(\theta)

To convert Cartesian coordinates to polar coordinate:

  • r=x2+y2r = \sqrt{x^2 + y^2}
  • θ=arctan(yx)\theta = \arctan(\frac{y}{x})

Polar graphs

The graph of an equation in polar coordinates is the set of all points PP whose polar coordinates satisfy the given equation. Solve by constructing a table of values for rr and solving for θ\theta.

Polar area

Similar to Riemann sums in rectangular form, instead of using rectangular areas as the basic units being summed, we sum areas of circular sectors.

The area of a circular sector if radius rr is given by: A=12r2θA = \frac{1}{2}r^2\theta

Using this formula, a formula for finding the area enclosed by a polar curve is:

A=ab12[f(θ)]2dθA = \int_{a}^{b} \frac{1}{2}[f(\theta)]^2 d\theta

Intersection of polar-form curves

  1. Find all simultaneous solutions of the given system of equations.
  2. Determine whether the pole r=0r=0 like on the two graphs.
  3. Graph the curves to look for other points of intersection.

6.4 Arc Length and Surface Area

Arc length

Arc length formula in integral form:

for y=f(x)y =f(x)

s=ab1+(dydx)2dxs = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx

for x=g(y)x =g(y)

s=cd1+(dgdy)2dys = \int_{c}^{d} \sqrt{1 + \left( \frac{dg}{dy} \right)^2} dy

Surface area

for y=f(x)y =f(x)

s=ab2πy1+(dydx)2dxs = \int_{a}^{b} 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx

6.5 Physical Applications


The work done by a constant force if a body moves a distance dd in the direction of an applied constant force FF, the work WW done is:


The work done by a variable force F(x)F(x) in moving an object along the xx-axis from x=ax=a to x=bx=b is given by:

W=abF(x)dxW=\int_{a}^{b} F(x) dx

Hooke’s law states that the force FF on a spring is proportional to the displacement xx:

F(x)=kxF(x) = kx

where kk is the spring constant.

To model work using Hooke’s formula, (1) solve for kk then (2) use the work done by a variable force formula to solve for Hooke’s formula using the displacement as the integral’s lower and upper limits.

F(x)=abkx dxF(x) = \int_{a}^{b} kx \ dx

If you are given work done WW instead of force FF, integration is required first to solve for kk

abkx dx=W\int_{a}^{b} kx \ dx = W

where aa and bb are the given displacement, then plug kk into the work done by a variable force formula as above.

Work done in emptying a tank

Force=Density×Volume\text{Force} = \text{Density} \times \text{Volume}

First, set up a coordinate system; decide where to place x=0x = 0.

To calculate work done in emptying a tank, use the work done by a variable force equation where the lower limit aa is the minimum x-value of the liquid and the upper limit bb is the maximum x-value of the liquid. Remember that F(x)=dFF(x) = dF where dd is the distance between xx and its destination (largersmaller)(\text{larger} - \text{smaller}) and force FF is the volume of xx, a “slice” of liquid vxv_{x} times density ρ\rho i.e. (πr2dx)(ρ)(\pi r^2 dx)(\rho).

abd(vxρ)\int_{a}^{b} d(v_{x} \rho)

Note that the volume of a “slice” depends on the shape of the tank but always uses dxdx as a dimension. For example a “slice” in a cylindrical tank will have a volume of πr2dx\pi r^2 dx.

Hydrostatic force problems

Force=Density×Depth×Area\text{Force} = \text{Density} \times \text{Depth} \times \text{Area}

First, set up a coordinate system i.e. let x=0x = 0 be the waterline. Then let Depth\text{Depth} equal the distance between xx and its destination (largersmaller)(\text{larger} - \text{smaller}). Remember that Area\text{Area} is the area of xx (a “slice”) and not the total surface area. Integrate from xminx_{min} to xmaxx_{max} of the fluid.

Moments and center of mass

xˉ\bar{x} is the center of mass: the point where the fulcrum should be placed to make the system balance.

Moments measure the tendency of a body to rotate about an axis.

Let ρ\rho denote density of the lamina.

  1. The mass of the lamina is
m=ρabf(x)dxm = \rho \int_{a}^{b} f(x) dx
  1. The moment MxM_{x} with respect to the xx-axis is
Mx=ab(f(x)+g(x)2)(f(x)g(x))dxM_{x} = \int_{a}^{b} \left( \frac{f(x) + g(x)}{2} \right)(f(x) - g(x))dx
  1. The moment MyM_{y} with respect to the yy-axis is
My=abx(f(x)g(x))dxM_{y} = \int_{a}^{b} x(f(x) - g(x))dx
  1. The coordinates of the center of mass (xˉ,yˉ)(\bar{x}, \bar{y}) of the system are
xˉ=Mym\bar{x} = \frac{M_{y}}{m}
yˉ=Mxm\bar{y} = \frac{M_{x}}{m}

where MxM_{x} is moment about the xx-axis and MyM_{y} is moment about the yy-axis.

Finding the centroid of a region bounded by two functions

  1. Calculate the points of intersection to set the limits of integration

Solve for f(x)=g(x)f(x) = g(x)

Then set ab\int_{a}^{b}

  1. Calculate the total mass

Assuming we are calculating for a lamina and ρ=1\rho = 1 (See area between two curves)

ab[f(x)g(x)]dx\int_{a}^{b} [f(x) - g(x)]dx

where f(x)g(x)f(x) \geq g(x) on [a,b][a, b].

  1. Compute the moments

Distance to yy-axis is xx; distance to xx-axis is the average of f(x)f(x) and g(x)g(x).

where f(x)g(x)f(x) \geq g(x) on [a,b][a, b].

  1. Solve for the coordinates of the center of mass.

7 Methods of Integration

7.1 Integration by Parts

Integration by parts is the inverse of the product rule for differentiation: d(uv)=u dv+v dud(uv) = u \ dv + v \ du.

Integrate both sides to find the formula for integration by parts: uv=u dv+v duuv = \int u \ dv + \int v \ du.

Rewritten as

u dv=uvv du\int u \ dv = uv - \int v \ du

Integration by parts using a table

If the derivative of uu is eventually 00, you can use a table to obtain the answer.

Example: x3sin(x)dx\int x^3 \sin(x)dx

Derivatives Integrals
x3x^3 sin(x)\sin(x)
3x23x^2 cos(x)-\cos(x)
6x6x sin(x)-\sin(x)
66 cos(x)\cos(x)
00 ssin(x)s\sin(x)

Use the table to keep track of each “round’s” uvuv (the next integral in the table).

Answer: x3(cosx)3x2(sinx)+6x(cosx)6(sinx)+Cx^3(-\cos x) - 3x^2(-\sin x) + 6x(\cos x) - 6(\sin x) + C

7.3 Trigonometric Methods

Powers of sine and cosine

sinmxcosnxdx\int \sin^m x \cos^n x dx

For even powers, square the trig identity.

For odd powers, peel off one trig function to become dudu.

Powers of secant and tangent

7.4 Method of Partial Fractions

This process can be thought of as the reverse of adding fractional algebraic expressions, and it allows us to break up rational expressions into simpler terms.

General rules:

  • Linear terms get constants AA
  • Quadratic terms get linear terms Bx+CBx + C
  • Multiple terms get repeated

7.7 Improper Integrals

Improper integrals with unbounded endpoints

From finite to \infty

1f(x)dx=limN1Nf(x)dx\int_{1}^{\infty} f(x) dx = \lim\limits_{N \to \infty} \int_{1}^{N} f(x) dx

If the limit is finite, we say that the improper integral converges, otherwise, the integral diverges.


11xpdxconvergesp>1\int_{1}^{\infty} \frac{1}{x^p} dx \hspace{1em} \text{converges} \Leftrightarrow p > 1

From -\infty to \infty

f(x)dx=0f(x)dx+0f(x)dx\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{0} f(x) dx + \int_{0}^{\infty} f(x) dx

Improper integrals with unbounded functions

7.8 Hyperbolic and Inverse Hyperbolic Functions

ddxcoshx=sinhx\frac{d}{dx} \cosh x = \sinh x
ddxsinhx=coshx\frac{d}{dx} \sinh x = \cosh x

8 Infinite Series

8.1 Sequences and their limits


A sequence is a function whose domain is N\N. The functional values a1,a2,a3,...a_1, a_2, a_3, ... are called the terms of the sequence, and ana_n is called the nnth term, or general term, of the sequence written as

(Sn)or{Sn}(S_n) \hspace{1em} \text{or} \hspace{1em} \{S_n\}

Limits of sequences

Growth (quickest to slowest)

If the bottom of a limit is more “powerful,” the fraction goes to 00.: limnnn2=0\lim\limits_{n \to \infty} \frac{n}{n^2} = 0

If the top of a limit is more “powerful,” the fraction goes to \infty: limnn2n=\lim\limits_{n \to \infty} \frac{n^2}{n} = \infty

If the top and bottom are equally “powerful,” break out the coefficients: limn2n3n=23\lim\limits_{n \to \infty} \frac{2n}{3n} = \frac{2}{3}

Facts to know

limn(1+1k)k=e\lim\limits_{n \to \infty} \left(1 + \frac{1}{k} \right)^k = e
limnk1k=1\lim\limits_{n \to \infty} k^{\frac{1}{k}} = 1
limnak=1\lim\limits_{n \to \infty} \sqrt[k]{a} = 1

Bounded, monotonic sequences

Name Condition
Strictly increasing an+1>ana_{n + 1} > a_n for all nn
Increasing an+1ana_{n + 1} \geq a_n for all nn
Strictly decreasing an+1<ana_{n + 1} < a_n for all nn
Decreasing an+1ana_{n + 1} \leq a_n for all nn
Monotone if either increasing or decreasing
Bounded above by MM anMa_n \leq M for all nn
Bounded below by mm anma_n \geq m for all nn
Bounded if both bounded above and below

8.2 Introduction to infinite series; Geometric series

A series is the sum of a sequence; an expression of the form

a1+a2+a3+...=k=1aka_1 + a_2 + a_3 + ... = \sum_{k=1}^\infty a_k

and the nnth partial sum of the series is

Sn=a1+a2+...+an=k=1nakS_n = a_1 + a_2 + ... + a_n = \sum_{k=1}^n a_k

The series converges if the sequence of partial sums converges.

A geometric series

arn+arn+1+...=k=nark{converges toarn1rifr<1diverges ifr1 ar^{n} + ar^{n+1} + ... = \sum_{k=n}^\infty ar^k \left\{ \begin{array}{l} \text{converges to} \hspace{1em} \frac{ar^n}{1 - r} \hspace{1em} \text{if} \hspace{1em} |r| < 1\\ \text{diverges if} \hspace{1em} |r| \geq 1 \end{array} \right.

where aa is the first term, rr is the common ratio, and nn is the number of terms.

For telescoping series, in the form

k=11k(k+1)\sum_{k=1}^\infty \frac{1}{k(k + 1)}

use partial fractions.

k=11k1k+1\sum_{k=1}^\infty \frac{1}{k} - \frac{1}{k + 1}

Notice how part of the terms cancel out.

Sn=(112)+(1213)+(1314)+...+(1n1n+1)=11n+1S_n = \left(1 - \cancel{\frac{1}{2}}\right) + \left( \cancel{\frac{1}{2}} - \cancel{\frac{1}{3}} \right) + \left( \cancel{\frac{1}{3}} - \cancel{\frac{1}{4}} \right) + ... + \left( \cancel{\frac{1}{n}} - \frac{1}{n + 1} \right) = 1 - \frac{1}{n + 1}


limnSn=limn11n+1=1\lim\limits_{n \to \infty} S_n = \lim\limits_{n \to \infty} 1 - \frac{1}{n + 1} = 1

8.3 The Integral Test; p-series

The convergence or divergence of an infinite series is determined by the behavior of its nnth partial sum, SnS_n, as nn \to \infty. Unfortunately, it is often difficult or impossible to find a usable formula for the nnth partial sum of a series, and other techniques must be used.

Divergence test

Theorem: The divergence test. If limk0\lim\limits_{k \to \infty} \neq 0, then the series Σak\Sigma a_k must diverge.

Warning: You can never say a series converges because of the divergence test.

Integral test; Series of non-negative numbers

Theorem: Convergence criterion for series with non-negative terms. A series Σak0\Sigma a_k \geq 0 for all kk converges if its sequence of partial sums is bounded from above and diverges otherwise.

Theorem: The integral test. If ak=f(x)a_k = f(x) for k=1,2,...k = 1, 2, ..., where ff is a positive, continuous, and decreasing function of xx for x1x \leq 1, then

k=1akand1f(x)dx\sum_{k=1}^\infty a_k \hspace{1em} \text{and} \hspace{1em} \int_{1}^{\infty} f(x) dx

either both converge or both diverge.

The harmonic series

k=11k\sum_{k=1}^\infty \frac{1}{k}

diverges by the integral test.

11xdx=limb1b1xdx=limb[lnbln1]=\int_{1}^\infty \frac{1}{x} dx = \lim\limits_{b \to \infty} \int_{1}^{b} \frac{1}{x} dx = \lim\limits_{b \to \infty} [\ln b - \ln 1] = \infty

P-series test

The harmonic series is a special case of a more general series called a p-series.

k=11kp\sum_{k=1}^\infty \frac{1}{k^p}

where pp is a positive constant.

Theorem: The p-series test. The p-series k=11kp\sum_{k=1}^\infty \frac{1}{k^p} converges if p>1p > 1 and diverges if p1p \leq 1.

8.4 Comparison Tests

Direct comparison test

Theorem: Direct comparison test. Suppose 0akbk0 \leq a_k \leq b_k for all kk.

If Σbk\Sigma b_k converges, then Σak\Sigma a_k converges.

If Σak\Sigma a_k diverges, then Σbk\Sigma b_k diverges.

Limit comparison test

Theorem: Limit comparison test. Suppose ak>0a_k > 0 and ab>0a_b > 0 for all sufficiently large kk.

limkakbk=L\lim\limits_{k \to \infty} \frac{a_k}{b_k} = L

where LL is finite and positive (is not zero or infinity). Then Σak\Sigma a_k and Σbk\Sigma b_k either both converge or both diverge.

Zero-infinity comparison test

Theorem: Zero-infinity limit comparison test. Suppose ak>0a_k > 0 and ab>0a_b > 0 for all sufficiently large kk.

If limkakbk=0\lim\limits_{k \to \infty} \frac{a_k}{b_k} = 0 and Σbk\Sigma b_k converges, then the series Σak\Sigma a_k converges.

If limkakbk=\lim\limits_{k \to \infty} \frac{a_k}{b_k} = \infty and Σbk\Sigma b_k diverges, then the series Σak\Sigma a_k diverges.

8.5 The Ratio Test and the Root Test

Ratio test

Theorem: The ratio test. Given the series Σak\Sigma a_k with ak>0a_k > 0, and limkak+1ak=L\lim\limits_{k \to \infty} \frac{a_{k + 1}}{a_k} = L

The ratio test states the following:

  • If L<1L < 1, then Σak\Sigma a_k converges.
  • If L>1L > 1, or if LL is infinite, then Σak\Sigma a_k diverges.
  • If L=1L = 1, then the test is inconclusive.

Root test

Theorem: The root test. Given the series Σak\Sigma a_k with ak>0a_k > 0, and limkakk=L\lim\limits_{k \to \infty} \sqrt[k]{a_k} = L

The root test states the following:

  • If L<1L < 1, then Σak\Sigma a_k converges.
  • If L>1L > 1, or if LL is infinite, then Σak\Sigma a_k diverges.
  • If L=1L = 1, then the test is inconclusive.

8.6 Alternating Series; Absolute and Conditional Convergence

Alternating series test

Theorem: The alternating series test. An alternating series

(1)kakor(1)k+1ak\sum (-1)^k a_k \hspace{1em} \text{or} \hspace{1em} \sum (-1)^{k+1} a_k

with ak>0a_k > 0 converges if both

  1. limkak=0\lim\limits_{k \to \infty} a_k = 0
  2. {ak}\{a_k\} is a decreasing sequence

Absolute convergence test

Theorem: The absolute convergence test. A series of real numbers Σak\Sigma a_k must converge if the related absolute value series Σak\Sigma |a_k| converges.

8.7 Power Series

8.8 Taylor and Maclaurin Series