CAS MA 124 Study Guide - Midterm Guide: Antiderivative, Improper Integral, Indeterminate Form
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10 Oct 2018
School
Department
Course
Professor
MA 124 CALCULUS II C1, Solutions to First Midterm Exam
Prof. Nikola Popovic, February 16, 2006, 08:00am - 09:20am
Problem 1 (15 points).
Determine whether the statements below are true or false. If a statement is true, explain why; if it
is false, give a counter-example.
(a) If fand gare continuous on [a, b], then
Zb
a
[f(x)·g(x)]dx =Zb
a
f(x)dx·Zb
a
g(x)dx.
(b) If fis a continuous, decreasing function on [a, ∞) and limx→∞ f(x)= 0, then Z∞
a
f(x)dx
is convergent.
(c) All continuous functions have antiderivatives.
Solution.
(a) FALSE. Take e.g. f(x)=x,g(x)=x, and [a, b] = [0,1]. Then,
Zb
a
[f(x)·g(x)]dx =Z1
0
x·x dx =Z1
0
x2dx =x3
3
1
0
=1
3,
but
Zb
a
f(x)dx·Zb
a
g(x)dx=Z1
0
x dx·Z1
0
x dx=x2
2
1
0·x2
2
1
0
=1
2·1
2=1
4.
(b) FALSE. Take e.g. f(x)=1
xand a= 1; then, f(x) is continuous on [1,∞), limx→∞ f(x)= 0,
and f(x) is decreasing on [1,∞) (since f0(x)=−1
x2<0), but
Z∞
1
1
xdx =lim
t→∞ Zt
1
1
xdx =lim
t→∞ ln x
t
1
=lim
t→∞ ln t=∞.
Hence, the improper integral diverges.
(c) TRUE. By the Fundamental Theorem of Calculus, Part I, we know that if f(x) is continuous
on [a, b],
g(x)=Zx
a
f(t)dt
is an antiderivative of f, i.e., g0(x)=f(x). (Note that this does not necessarily imply that we
can compute gexplicitly; take e.g. f(x)= ex2. However, even in that case, we know that g
exists.)