MATH 142 Exam with 100% Verified
Solutions Exam Overview
MATH 142 covers integral calculus, including techniques of integration
(substitution, integration by parts, partial fractions, trigonometric
substitution), applications of integration (area, volume, work, arc length,
average value), and sequences and series. This practice exam provides
comprehensive coverage of the topics typically tested in MATH 142 exams.
SECTION 1: INTEGRATION TECHNIQUES
1. Evaluate \int \frac{e^x}{\sqrt{1-e^x}} dx∫
1−e
x
e
x
dx
Solution:
Let u = 1 - e^xu=1−e
x
. Then du = -e^x dxdu=−e
x
dx, so e^x dx = -due
x
dx=−du.
,\int \frac{e^x}{\sqrt{1-e^x}} dx = -\int u^{-1/2} du = -2\sqrt{u} + C =
\boxed{-2\sqrt{1-e^x} + C}∫
1−e
x
e
x
dx=−∫u
−1/2
du=−2
u
+C=
−2
1−e
x
+C
Rationale: Substitution is the appropriate technique when the derivative of
the inner function appears in the integrand. Notice that e^xe
x
is present, and its derivative also appears.
2. Evaluate \int (2^x + x^2) dx∫(2
x
+x
,2
)dx
Solution:
\int 2^x dx = \int e^{x\ln 2} dx = \frac{1}{\ln 2} 2^x∫2
x
dx=∫e
xln2
dx=
ln2
1
2
x
Thus,
\int (2^x + x^2) dx = \boxed{\frac{2^x}{\ln 2} + \frac{x^3}{3} + C}∫(2
x
+x
2
)dx=
ln2
2
x
+
3
x
3
, +C
Rationale: For exponential functions with bases other than e, use the
identity a^x = e^{x\ln a}a
x
=e
xlna
.
3. Evaluate \int_0^{\frac{\sqrt{3}}{3}} \frac{2x}{3x^2+1} dx∫
0
3
3
3x
2
+1
2x
dx
Solution:
Let u = 3x^2 + 1u=3x
2
+1, du = 6x dxdu=6xdx, so 2x dx = \frac{du}{3}2xdx=
3
du
Solutions Exam Overview
MATH 142 covers integral calculus, including techniques of integration
(substitution, integration by parts, partial fractions, trigonometric
substitution), applications of integration (area, volume, work, arc length,
average value), and sequences and series. This practice exam provides
comprehensive coverage of the topics typically tested in MATH 142 exams.
SECTION 1: INTEGRATION TECHNIQUES
1. Evaluate \int \frac{e^x}{\sqrt{1-e^x}} dx∫
1−e
x
e
x
dx
Solution:
Let u = 1 - e^xu=1−e
x
. Then du = -e^x dxdu=−e
x
dx, so e^x dx = -due
x
dx=−du.
,\int \frac{e^x}{\sqrt{1-e^x}} dx = -\int u^{-1/2} du = -2\sqrt{u} + C =
\boxed{-2\sqrt{1-e^x} + C}∫
1−e
x
e
x
dx=−∫u
−1/2
du=−2
u
+C=
−2
1−e
x
+C
Rationale: Substitution is the appropriate technique when the derivative of
the inner function appears in the integrand. Notice that e^xe
x
is present, and its derivative also appears.
2. Evaluate \int (2^x + x^2) dx∫(2
x
+x
,2
)dx
Solution:
\int 2^x dx = \int e^{x\ln 2} dx = \frac{1}{\ln 2} 2^x∫2
x
dx=∫e
xln2
dx=
ln2
1
2
x
Thus,
\int (2^x + x^2) dx = \boxed{\frac{2^x}{\ln 2} + \frac{x^3}{3} + C}∫(2
x
+x
2
)dx=
ln2
2
x
+
3
x
3
, +C
Rationale: For exponential functions with bases other than e, use the
identity a^x = e^{x\ln a}a
x
=e
xlna
.
3. Evaluate \int_0^{\frac{\sqrt{3}}{3}} \frac{2x}{3x^2+1} dx∫
0
3
3
3x
2
+1
2x
dx
Solution:
Let u = 3x^2 + 1u=3x
2
+1, du = 6x dxdu=6xdx, so 2x dx = \frac{du}{3}2xdx=
3
du
