Calculus 2

1
By Gökhan Bilhan
Calculus 2
(Week 6)-Improper Integrals Part 1
How can we nd the area under the curve, or in other words, what is the result of the integral
Z
∞
1
lnx
dx
x2
Denition. Let's consider the integral
∞
Z
f (x)dx.
a
Then we do:
Z
θ
f (x)dx
limθ7→∞
a
If the case is
Z
b
f (x)dx then we do:
−∞
Z
limθ7→−∞
b
f (x)dx
θ
If the case is
Z
Z
−∞
−∞
Z c
∞
f (x)dx =
−∞
−∞
f (x)dx then we split the integral into two parts at any point c:
Z ∞
f (x)dx +
f (x)dx and then we go on.
c
Note that, if the result of the limit is a number then our integral's result is this number and we
say that the integral is CONVERGENT.
If the result of the limit is +∞ or −∞, then we say that the integral is DIVERGENT.
Example: Now, let's calculate the integral
Z
1
∞
lnx
dx
x2
2
By Gökhan Bilhan
..
Example:
Z
∞
−∞
1
dx
1 + x2
3
By Gökhan Bilhan
Example: For what values of "p" does the integral
Z
1
p>1
p<1
p=1
∞
1
dx converge, what is its value?
xp
4
By Gökhan Bilhan
Exercises
1. Determine whether the following integrals are convergent or not. In case an integral is convergent, evaluate it
Z
∞
1
Z
∞
∞
1
∞
Z
∞
1
dx
x
Z
∞
1
dx
Z
dx
0,1001
1
√ dx
x
∞
1
Z
1
1
x1,0000000000001
1
1
x
1
Z
1
x1001
1
Z
1
dx
x2
∞
dx
1
dx
xπ
1
√ dx
x
1001
2. Evaluate the integral or state that it diverges.
Z
0
∞
(1 +
x2 )(1
1
dx
+ arctanx)
5
By Gökhan Bilhan
3. Evaluate the integral or state that it diverges.
Z
2
−∞
x2
2
dx
+4
6
By Gökhan Bilhan
PART 2
Denition. Let's consider the integral
b
Z
f (x)dx.
a
θ.
If at any of the boundaries, our function y = f (x) becomes not continuous, then we replace it with
In particular, say f (a) is undened, i.e. at x = a, the function y = f (x) becomes discontinuous,
then we do:
Z
b
limθ7→a+
f (x)dx
θ
If say f (b) is undened, i.e. at x = b, the function y = f (x) becomes discontinuous, then we do:
Z
θ
f (x)dx
limθ7→b−
a
Example: Let's investigate the convergence of
Z
0
1
1
dx
1−x
7
By Gökhan Bilhan
Denition. Let's consider the integral
Z
b
f (x)dx.
a
If at a point c among the boundaries a and b, our function y = f (x) becomes not continuous,
then we split it into two parts at the point c and then apply the method described at the preceeding
denition.
i.e.
Z
b
Z
Z
a
b
f (x)dx
f (x)dx +
f (x)dx =
a
c
c
Example: Let's investigate the convergence of
Z
3
1
2
0
(x − 1) 3
dx
8
By Gökhan Bilhan
Exercises
1. For which values of p, the integral
Z
0
1
1
dx is convergent?
xp
2. Evaluate the integral or state that it diverges.
Z
1
3.
Z
0
∞
1
dx
x2
2
1
ds
s s2 − 1
√
9
By Gökhan Bilhan
4. Evaluate the integral or state that it diverges.
Z
0
1
√
4r
dr
1 − r4
By Gökhan Bilhan
Various Exercises
1.
2.
1
1
−1
x3
Z
2
Z
2
s+1
√
ds
4 − s2
4
1
p dx
|x|
0
3.
Z
−1
4.
Z
dx
π
2
cotxdx
0
10
11
By Gökhan Bilhan
5.
Z
∞
√
4
6.
Z
∞
√
1
7.
Z
2
∞
√
1
dx
x−1
x+1
dx
x2
1
x2 − 1
dx