Algebra applications of integration

CHAPTER 2

Applications of Integration


2.1. More about Areas

2.1.1. Area Between Two Curves. The area between the curves
y = f (x) and y = g(x) and the lines x = a and x = b (f , g continuous
and f (x) ≥ g(x) for x in [a, b]) is
Z b Z b Z b
A= f (x) dx − g(x) dx = [f (x) − g(x)] dx .
a a a

Calling yT = f (x), yB = g(x), we have:
Z b
A= (yT − yB ) dx
a


Example: Find the area between y = ex and y = x bounded on the
sides by x = 0 and x = 1.
Answer : First note that ex ≥ x for 0 ≤ x ≤ 1. So:
Z 1 · ¸1 µ ¶ µ ¶
x x x2 1 12 0 02
A= (e − x) dx = e − = e − − e −
0 2 0 2 2

= e − 12 − 1 = e − 3
2
.


The area between two curves y = f (x) and y = g(x) that intersect
at two points can be computed in the following way. First find the
intersection points a and b by solving the equation f (x) = g(x). Then
find the difference:
Z b Z b Z b
f (x) dx − g(x) dx = [f (x) − g(x)] dx .
a a a

If the result is negative that means that we have subtracted wrong.
Just take the result in absolute value.
50

, 2.1. MORE ABOUT AREAS 51

Example: Find the area between y = x2 and y = 2 − x. Solution:
First, find the intersection points by solving x2 −(2−x) = x2 +x−2 = 0.
We get x = −2 and x = 1. Next compute:
Z 1 Z 1
2
(x − (2 − x)) dx = (x2 + x − 2) dx = −9/2 .
−2 −2
Hence the area is 9/2.
Sometimes it is easier or more convenient to write x as a function of
y and integrate respect to y. If xL (y) ≤ xR (y) for p ≤ y ≤ q, then the
area between the graphs of x = xL (y) and x = xR (y) and the horizontal
lines y = p and y = q is:
Z q
A= (xR − xL ) dy
p


Example: Find the area between the line y = x−1 and the parabola
y 2 = 2x + 6.

4




2




–2 0 2 4 6
x


–2




–4




Answer : The intersection points between those curves are (−1, −2)
and (5, 4), but in the figure we can see that the region extends to the
left of x = −1. In this case it is easier to write
xL = 12 y 2 − 3 , xR = y + 1 ,
and integrate from y = −2 to y = 4:
Z 4 Z 4
© ª
A= (xR − xL ) dx = (y + 1) − ( 12 y 2 − 3) dx
−2 −2
Z 4 ¡ 1 2 ¢
= − 2 y + y + 4 dx
−2
· ¸4
y3 y2
= − + + 4y
6 2 −2

= 18

, 2.2. VOLUMES 52

2.2. Volumes

2.2.1. Volumes by Slices. First we study how to find the volume
of some solids by the method of cross sections (or “slices”). The idea
is to divide the solid into slices perpendicular to a given reference line.
The volume of the solid is the sum of the volumes of its slices.

2.2.2. Volume of Cylinders. A cylinder is a solid whose cross
sections are parallel translations of one another. The volume of a cylin-
der is the product of its height and the area of its base:
V = Ah .

2.2.3. Volume by Cross Sections. Let R be a solid lying along-
side some interval [a, b] of the x-axis. For each x in [a, b] we denote A(x)
the area of the cross section of the solid by a plane perpendicular to
the x-axis at x. We divide the interval into n subintervals [xi−1 , xi ], of
length ∆x = (b − a)/n each. The planes that are perpendicular to the
x-axis at the points x0 , x1 , x2 , . . . , xn divide the solid into n slices. If
the cross section of R changes little along a subinterval [xi−1 , xi ], the
slab positioned alongside that subinterval can be considered a cylinder
of height ∆x and whose base equals the cross section A(x∗i ) at some
point x∗i in [xi−1 , xi ]. So the volume of the slice is
∆Vi ≈ A(x∗i ) ∆x .
The total volume of the solid is
Xn X
n
V = ∆Vi ≈ A(x∗i ) ∆x .
i=1 i=1

Once again we recognize a Riemann sum at the right. In the limit as
n → ∞ we get the so called Cavalieri’s principle:
Z b
V = A(x) dx .
a


Of course, the formula can be applied to any axis. For instance
if a solid lies alongside some interval [a, b] on the y axis, the formula
becomes Z b
V = A(y) dy .
a

Example: Find the volume of a cone of radius r and height h.