Worksheet 5
Spring 2016
MATH 222, Week 5: 2.5,3.1,3.2,3.3
Name:
You aren’t necessarily expected to finish the entire worksheet in discussion. There are a lot of problems to
supplement your homework and general problem bank for studying.
Determine the convergence or divergence of the following integrals without computing them. Then
compute them explicitly.
Z
∞
Problem 1.
4
1
dx.
x−2
Solution 1.
1
> x1 . We don’t have to worry about negatives because of the
Notice that x − 2 < x. This implies that x−2
interval of integration. The comparison test tells us that this integral diverges because x1 does.
To see this explicitly, we set up the limit and antidifferentiate:
a
Z a
1
lim
dx = lim ln |x − 2| = ∞
a→∞ 4 x − 2
a→∞
4
Z
∞
Problem 2.
4
1
dx.
x3 − x
Solution 2.
We want to compare this to something. Intuitively we would guess this behaves like x3 as x gets really big
so this should converge. We’ll use the limit comparison test with x13
1/x3
x3 − x
=
lim
=1
x→∞ 1/(x3 − x)
x→∞
x3
lim
As the limit is greater than zero and is constant we know that convergence of one integral is equivalent to convergence of the other. Thus our integral should converge.
To see this explicitly we’ll use partial fractions to integrate.
a
a
Z a
1
1
1
1
2 2
2 lim
= lim − ln |x| + ln |1 − x | = lim ln |(1 − x )/x | = ln(4) + ln(15)
a→∞ 4 x(x − 1)(x + 1)
a→∞
a→∞ 2
2
2
4
4
Z
Problem 3.
1
∞
dt
.
1 + e2x
Solution 3.
1
We expect this to converge as e2 x becomes massive as x → ∞. We know that e2x
converges. We know 1+e2x > e2x
1
1
this implies that 1+e2x < e2x . So by the comparison test our integral will converge.
To see this explicitly we can make a u-sub. Let u = ex , du = ex . Then our integral becomes:
∗
2x a
2 Z ∗
e
e
1
2 du = lim ln |u| − (1/2) ln |1 + u | = lim (1/2) ln
= −(1/2) ln
lim
2x
a→∞
a→∞
a→∞ ∗ u(1 + u2 )
1+e
1 + e2
∗
1
∞
sin(x2 )
dx. Don’t worry about computing this explicitly. Sketch the function in the integrand
x2
0
though and look at what happens at x = 0?
Z
Problem 4.
Solution 4.
We know that −1 ≤ sin(x2 ) ≤ 1, and so
compute it explicitly:
sin(x2 )
x2
≤
1
.
x2
b
Z
lim lim
a→0 b→∞ a
By the comparison test our integral converges. To
sin(x2 )
dx
x2
You could use integration by parts to solve this integral. Letting u = sin(x2 ) and v 0 = 1/(x2 ). After one interation
you have:
Z
Z
2x cos(x2 )
2
2
2
sin(x )/x +
dx = − sin(x )/x + 2 cos(x2 )
x
So we have:
b Z b
Z
2
lim lim − sin(x )/x +
2 cos(x ) dx = 2
2
a→0 b→∞
a
a
∞
cos(x2 ) dx
0
To find this we recognize this as a Fresnel integral, and when integrated from 0 to infinity it equals
not expected to know this, but I’m including it just in case anyone is curious! So in all
Z ∞
p
sin(x2 )
dx
=
π/2
x2
0
2
p
π/2. You’re
Problem 5. Find a solution to the initial value problem:
dy
= e y x3
dx
With initial value y(0) = 0.
Problem 6. Find a solution to the initial value problem:
p
dy
= y y 2 − 1 cos(x)
dx
With initial value y(0) = 1.
Problem 7. Find the general solution to the differential equation
dy
= x2 + y 2 x2
dx
3