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Millersville University
Department of Mathematics
MATH 211, Calculus II
Please evaluate the following improper integrals.
∫ (^1)
− 1
x^2 /^3
dx
This integral is improper since the integrand possesses a discontinuity at x = 0 and − 1 < 0 < 1.
∫ (^1)
− 1
x^2 /^3
dx = lim M → 0 −
∫ (^) M
− 1
x^2 /^3
dx + lim R→ 0 +
∫ (^1)
R
x^2 /^3
dx
= lim M → 0 −
∫ (^) M
− 1
x − 2 / 3 dx + lim R→ 0 +
∫ (^1)
R
x − 2 / 3 dx
= lim M → 0 −
( 3 x 1 / 3
∣ ∣ ∣
M
− 1
)
( 3 x 1 / 3
∣ ∣ ∣
1
R
)
= lim M → 0 −
( 3 M 1 / 3
)
( 3 − 3 R 1 / 3
)
∫ (^2)
−∞
x^2 + 4
dx
This integral is improper since the lower limit of integration is −∞.
∫ (^2)
−∞
x^2 + 4
dx = lim R→−∞
∫ (^2)
R
x^2 + 4
dx
= lim R→−∞
(
tan − 1 x 2
∣ ∣ ∣ ∣
2
R
)
= lim R→−∞
( tan − 1 (1) − tan
− 1 R
)
= lim R→−∞
( π
4
− tan
− 1 R
)
π
4
− lim R→−∞
( tan
− 1 R
)
π
4
( −
π
2
)
3 π
4
∫ (^2)
0
x + 1 √ 4 − x^2
dx
This integral is improper since the integrand is discontinuous at x = 2.
∫ (^2)
0
x + 1 √ 4 − x^2
dx = lim R→ 2 −
∫ (^) R
0
x + 1 √ 4 − x^2
dx
= lim R→ 2 −
∫ (^) R
0
( x √ 4 − x^2
4 − x^2
)
dx
= lim R→ 2 −
(
−
4 − x^2 + sin − 1 x 2
∣ ∣ ∣ ∣
R
0
)
= lim R→ 2 −
( −
4 − R^2 + 2 + sin−^1
R
)
∫ (^) ∞
− 1
x^2 + 5x + 6
dx
This integral is improper since the upper limit is integration is infinite.
∫ (^) ∞
− 1
x^2 + 5x + 6
dx = lim R→∞
∫ (^) R
− 1
x^2 + 5x + 6
dx
= lim R→∞
∫ (^) R
− 1
(x + 2)(x + 3)
dx
We must use partial fraction expansion to rewrite the integrand.
(x + 2)(x + 3)
A
x + 2
B
x + 3
A(x + 3) + B(x + 2)
(x + 2)(x + 3)
1 = A(x + 3) + B(x + 2)
When x = −2 we see that A = 1 and when x = −3 we have B = −1.
lim R→∞
∫ (^) R
− 1
(x + 2)(x + 3)
dx = lim R→∞
∫ (^) R
− 1
( 1
x + 2
x + 3
) dx
= lim R→∞
( ln |x + 2| − ln |x + 3||
R − 1
)
= lim R→∞
(ln |R + 2| − ln |R + 3| − ln 1 + ln 2)
= lim R→∞
(
ln
|R + 2|
|R + 3|
)
= lim R→∞
( ln
∣ ∣ ∣ ∣
R + 2
R + 3
∣ ∣ ∣ ∣ + ln 2
)
= ln 1 + ln 2
= ln 2
= lim R→−∞
( −1 + e −R^2
)
( −e −M 2
)
∫ (^) ∞
0
16 tan − 1 x
1 + x^2
dx
This integral is improper since the upper limit of integration is infinite.
∫ (^) ∞
0
16 tan − 1 x
1 + x^2
dx = lim R→∞
∫ (^) R
0
16 tan − 1 x
1 + x^2
dx
We will integrate by substitution using
u = tan − 1 x and du =
1 + x^2
dx.
lim R→∞
∫ (^) R
0
16 tan − 1 x
1 + x^2
dx = lim R→∞
∫ (^) tan− (^1) R
0
16 u du
= lim R→∞
( 8 u^2
∣ ∣ ∣
tan−^1 R 0
)
= lim R→∞
( 8
[ tan − 1 R
] 2 )
[ π
2
] 2
= 2 π 2
∫ (^) ∞
0
2 e −x sin x dx
This integral is improper since the upper limit of integration is infinite. In order to find an antiderivative of the integrand, we must use integration by parts, twice. We make the following assignments.
u = 2 e −x v = − cos x
du = − 2 e −x dx dv = sin x dx
Therefore ∫ 2 e −x sin x dx = − 2 e −x cos x −
∫ 2 e −x cos x dx.
For the second integration by parts step we make the assignments:
u = 2 e −x v = sin x
du = − 2 e −x dx dv = cos x dx.
Now we have ∫ 2 e −x sin x dx = − 2 e −x cos x −
∫ 2 e −x cos x dx
= − 2 e−x^ cos x −
( 2 e−x^ sin x +
∫ 2 e−x^ sin x dx
)
= − 2 e−x^ cos x − 2 e−x^ sin x −
∫ 2 e−x^ sin x dx
∫ 2 e −x sin x dx = − 2 e −x cos x − 2 e −x sin x + C ∫ 2 e −x sin x dx = −e −x cos x − e −x sin x + C
Armed with this antiderivative we may return to the task of evaluating the improper integral.
∫ (^) ∞
0
2 e −x sin x dx = lim R→∞
∫ (^) R
0
2 e −x sin x dx
= lim R→∞
( −e −x cos x − e −x sin x
∣ ∣ ∣
R
0
)
= lim R→∞
( −e −R cos R − e −R sin R + 1
)
= 1 − lim R→∞
( e −R cos R + e −R sin R
)
= 1 − lim R→∞
( cos R
eR^
sin R
eR
)
Note that the last limit was evaluated by use of the Squeeze Theorem.
∫ (^1)
0
(− ln x) dx
This integral is improper since the integrand is discontinuous at x = 0.
∫ (^1)
0
(− ln x) dx = lim R→ 0 +
∫ (^1)
R
(− ln x) dx
= lim R→ 0 +
( −x ln x + x| 1 R
)
= lim R→ 0 +^
(1 − (−R ln R + R))
= lim R→ 0 +^
(1 + R ln R − R))
= 1 + lim R→ 0 +
(R ln R)
= 1 + lim R→ 0 +
ln R 1 R
