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 x-axis
Let f(x) and g(x) be continuous functions such that f(x)≥g(x) over an interval [a,b]. Let R denote the region bounded above by f(x), below by g(x), on the left by x=a, and on the right by x=b. The area of R is given by
A=∫ab[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=∫ab∣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)]dx
Area of regions, integrating along the y-axis
Let u(y) and v(y) be continuous functions such that u(y)≥v(y) over an interval [c,d] along the y-axis. A horizontal strip has a width of u(y)−v(y). Thus, the integration formula for area is
A=∫cd[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 x-axis.
V=∫abA(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)]2dx
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)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=∫ab2πxf(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 O, called the origin or pole and a fixed ray emanating from the origin called the polar axis. We associate with each point P in the plane an ordered pair of numbers P(r,θ), where r (radial coordinate) is the distance from O to P and θ (polar angle) is the angle measured from the polar axis.
Changing coordinates
To convert polar coordinates to Cartesian coordinates:
x=rcos(θ)
y=rsin(θ)
To convert Cartesian coordinates to polar coordinate:
r=x2+y2
θ=arctan(xy)
Polar graphs
The graph of an equation in polar coordinates is the set of all points P whose polar coordinates satisfy the given equation. Solve by constructing a table of values for r and solving for θ.
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 r is given by: A=21r2θ
Using this formula, a formula for finding the area enclosed by a polar curve is:
A=∫ab21[f(θ)]2dθ
Intersection of polar-form curves
Find all simultaneous solutions of the given system of equations.
Determine whether the pole r=0 like on the two graphs.
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)
s=∫ab1+(dxdy)2dx
for x=g(y)
s=∫cd1+(dydg)2dy
Surface area
for y=f(x)
s=∫ab2πy1+(dxdy)2dx
6.5 Physical Applications
Work
The work done by a constant force if a body moves a distance d in the direction of an applied constant force F, the work W done is:
W=Fd
The work done by a variable forceF(x) in moving an object along the x-axis from x=a to x=b is given by:
W=∫abF(x)dx
Hooke’s law states that the force F on a spring is proportional to the displacement x:
F(x)=kx
where k is the spring constant.
To model work using Hooke’s formula, (1) solve for k 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)=∫abkxdx
If you are given work done W instead of force F, integration is required first to solve for k
∫abkxdx=W
where a and b are the given displacement, then plug k into the work done by a variable force formula as above.
Work done in emptying a tank
Force=Density×Volume
First, set up a coordinate system; decide where to place x=0.
To calculate work done in emptying a tank, use the work done by a variable force equation where the lower limit a is the minimum x-value of the liquid and the upper limit b is the maximum x-value of the liquid. Remember that F(x)=dF where d is the distance between x and its destination (larger−smaller) and force F is the volume of x, a “slice” of liquid vx times density ρ i.e. (πr2dx)(ρ).
∫abd(vxρ)
Note that the volume of a “slice” depends on the shape of the tank but always uses dx as a dimension. For example a “slice” in a cylindrical tank will have a volume of πr2dx.
Hydrostatic force problems
Force=Density×Depth×Area
First, set up a coordinate system i.e. let x=0 be the waterline. Then let Depth equal the distance between x and its destination (larger−smaller). Remember that Area is the area of x (a “slice”) and not the total surface area. Integrate from xmin to xmax of the fluid.
Moments and center of mass
xˉ 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 denote density of the lamina.
The mass of the lamina is
m=ρ∫abf(x)dx
The moment Mx with respect to the x-axis is
Mx=∫ab(2f(x)+g(x))(f(x)−g(x))dx
The moment My with respect to the y-axis is
My=∫abx(f(x)−g(x))dx
The coordinates of the center of mass (xˉ,yˉ) of the system are
xˉ=mMy
yˉ=mMx
where Mx is moment about the x-axis and My is moment about the y-axis.
Finding the centroid of a region bounded by two functions
Calculate the points of intersection to set the limits of integration
Solve for f(x)=g(x)
Then set ∫ab
Calculate the total mass
Assuming we are calculating for a lamina and ρ=1 (See area between two curves)
∫ab[f(x)−g(x)]dx
where f(x)≥g(x) on [a,b].
Compute the moments
Distance to y-axis is x; distance to x-axis is the average of f(x) and g(x).
where f(x)≥g(x) on [a,b].
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)=udv+vdu.
Integrate both sides to find the formula for integration by parts: uv=∫udv+∫vdu.
Rewritten as
∫udv=uv−∫vdu
Integration by parts using a table
If the derivative of u is eventually 0, you can use a table to obtain the answer.
Example: ∫x3sin(x)dx
Derivatives
Integrals
x3
sin(x)
3x2
−cos(x)
6x
−sin(x)
6
cos(x)
0
ssin(x)
Use the table to keep track of each “round’s” uv (the next integral in the table).
Answer: x3(−cosx)−3x2(−sinx)+6x(cosx)−6(sinx)+C
7.3 Trigonometric Methods
Powers of sine and cosine
∫sinmxcosnxdx
For even powers, square the trig identity.
For odd powers, peel off one trig function to become du.
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 A
Quadratic terms get linear terms Bx+C
Multiple terms get repeated
7.7 Improper Integrals
Improper integrals with unbounded endpoints
From finite to ∞
∫1∞f(x)dx=N→∞lim∫1Nf(x)dx
If the limit is finite, we say that the improper integral converges, otherwise, the integral diverges.
Rule:
∫1∞xp1dxconverges⇔p>1
From −∞ to ∞
∫−∞∞f(x)dx=∫−∞0f(x)dx+∫0∞f(x)dx
Improper integrals with unbounded functions
7.8 Hyperbolic and Inverse Hyperbolic Functions
dxdcoshx=sinhx
dxdsinhx=coshx
8 Infinite Series
8.1 Sequences and their limits
Sequences
A sequence is a function whose domain is N. The functional values a1,a2,a3,... are called the terms of the sequence, and an is called the nth term, or general term, of the sequence written as
(Sn)or{Sn}
Limits of sequences
Growth (quickest to slowest)
nn
n!
3n
en
n4
n2
n
n
4n
lnn
If the bottom of a limit is more “powerful,” the fraction goes to 0.: n→∞limn2n=0
If the top of a limit is more “powerful,” the fraction goes to ∞: n→∞limnn2=∞
If the top and bottom are equally “powerful,” break out the coefficients: n→∞lim3n2n=32
Facts to know
n→∞lim(1+k1)k=e
n→∞limkk1=1
n→∞limka=1
Bounded, monotonic sequences
Name
Condition
Strictly increasing
an+1>an for all n
Increasing
an+1≥an for all n
Strictly decreasing
an+1<an for all n
Decreasing
an+1≤an for all n
Monotone
if either increasing or decreasing
Bounded above by M
an≤M for all n
Bounded below by m
an≥m for all n
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=1∑∞ak
and the nth partial sum of the series is
Sn=a1+a2+...+an=k=1∑nak
The series converges if the sequence of partial sums converges.
The convergence or divergence of an infinite series is determined by the behavior of its nth partial sum, Sn, as n→∞. Unfortunately, it is often difficult or impossible to find a usable formula for the nth partial sum of a series, and other techniques must be used.
Divergence test
Theorem: The divergence test. If k→∞lim=0, then the series Σak 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 Σak≥0 for all k converges if its sequence of partial sums is bounded from above and diverges otherwise.
Theorem: The integral test. If ak=f(x) for k=1,2,..., where f is a positive, continuous, and decreasing function of x for x≤1, then
k=1∑∞akand∫1∞f(x)dx
either both converge or both diverge.
The harmonic series
k=1∑∞k1
diverges by the integral test.
∫1∞x1dx=b→∞lim∫1bx1dx=b→∞lim[lnb−ln1]=∞
P-series test
The harmonic series is a special case of a more general series called a p-series.
k=1∑∞kp1
where p is a positive constant.
Theorem: The p-series test. The p-series ∑k=1∞kp1 converges if p>1 and diverges if p≤1.
8.4 Comparison Tests
Direct comparison test
Theorem: Direct comparison test. Suppose 0≤ak≤bk for all k.
If Σbk converges, then Σak converges.
If Σak diverges, then Σbk diverges.
Limit comparison test
Theorem: Limit comparison test. Suppose ak>0 and ab>0 for all sufficiently large k.
k→∞limbkak=L
where L is finite and positive (is not zero or infinity). Then Σak and Σbk either both converge or both diverge.
Zero-infinity comparison test
Theorem: Zero-infinity limit comparison test. Suppose ak>0 and ab>0 for all sufficiently large k.
If k→∞limbkak=0 and Σbk converges, then the series Σak converges.
If k→∞limbkak=∞ and Σbk diverges, then the series Σak diverges.
8.5 The Ratio Test and the Root Test
Ratio test
Theorem: The ratio test. Given the series Σak with ak>0, and k→∞limakak+1=L
The ratio test states the following:
If L<1, then Σak converges.
If L>1, or if L is infinite, then Σak diverges.
If L=1, then the test is inconclusive.
Root test
Theorem: The root test. Given the series Σak with ak>0, and k→∞limkak=L
The root test states the following:
If L<1, then Σak converges.
If L>1, or if L is infinite, then Σak diverges.
If L=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
with ak>0 converges if both
k→∞limak=0
{ak} is a decreasing sequence
Absolute convergence test
Theorem: The absolute convergence test. A series of real numbers Σak must converge if the related absolute value series Σ∣ak∣ converges.