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KOC¸ UNIVERSITY
MATH 106
FINAL EXAM MAY 26, 2012
Duration of Exam: 120 minutes
INSTRUCTIONS: No calculators may be used on the test. No questions, and talking
allowed. You must always explain your answers and show your work to receive full
credit. Use the back of these pages if necessary. Print (use CAPITAL LETTERS) and
sign your name. GOOD LUCK!
Solutions by Ali Alp Uzman and Candan G¨ud¨uc¨u
(Check One):
(Emre Alkan) : —–
(Burak Ozbaˇ¨ gcı): —–
PROBLEM POINTS SCORE
1 10
2 15
3 15
4 15
5 15
6 20
7 10
TOTAL 100
Problem 1 (10 pts) Find the limit
lim x→ 2
x
x − 2
∫ (^) x
2
cos t
t
dt.
SOLUTION.
lim x→ 2
x
x − 2
∫ (^) x
2
cos t
t
dt = lim x→ 2
x
∫ (^) x
2
cos t t
dt
x − 2
This is a
0 0
indeterminate form. So using L’Hospital’s Rule, product rule of deriva-
tives and the Fundamental Theorem of Calculus we get
lim x→ 2
x
∫ (^) x
2
cos t t
dt
x − 2
= lim x→ 2
x
cos x
x
∫ (^) x
2
cos t
t
dt = cos 2.
Problem 2 (15 pts) Find the number b such that the line y = b divides the region
bounded by the curves y = x
2 and y = 4 into two regions with equal area.
SOLUTION.
∫ (^) b
0
ydy =
b
ydy
y
3 / 2
b
0
y
3 / 2
4
b
y
3 / 2
b
0
= y
3 / 2
4
b
b
3 / 2 = 4
3 / 2 − b
3 / 2
b
3 / 2
3 / 2
b = 2
3
Problem 3 (15 pts) Using integration, find the volume of a pyramid with height
h and square base with side length a.
SOLUTION. The area of the cross section parallel to the base of the pyramid
whose distance from top vertex equals y is given by
a
2 y
2
h^2
. Hence the volume is
V =
∫ (^) h
0
A(y)dy =
∫ (^) h
0
a
2 y
2
h^2
dy =
a
2 h
Problem 6 (20 pts) Decide whether the improper integral
∫ (^) ∞
2
x
x^2 − 4
dx
is convergent or divergent. If it is convergent, then find the exact value.
SOLUTION
This integral is improper for two reason;
(i)[2, ∞) is infinite,
(ii) The integrand has an infinite discontinuity at 2.
2
x
x^2 − 4
dx =
2
x
x^2 − 4
dx +
3
x
x^2 − 4
dx
Using substitution
x^2 − 4 = 2t , x = 2
t^2 + 1 and dx = √^2 t t^2 +
dt
x
x
2 − 4
dx =
2(t^2 + 1)
dt =
arctan t
arctan(
√ x^2 − 4 2
2
x
x^2 − 4
dx = lim a→ 2
a
x
x^2 − 4
dx = lim a→ 2
arctan(
√ x^2 − 4 2
3
a
= lim a→ 2
arctan(
√ 5 2
arctan(
√ a^2 − 4 2
arctan(
√ 5 2
arctan 0
arctan(
√ 5 2
3
x
x^2 − 4
dx = lim b→∞
∫ (^) b
3
x
x^2 − 4
dx = lim b→∞
arctan(
√ x^2 − 4 2
b
3
= lim b→∞
arctan(
√ b^2 − 4 2
arctan(
√ 5 2
π
arctan(
√ 5 2
2
x
x
2 − 4
dx =
2
x
x
2 − 4
dx +
3
x
x
2 − 4
dx
π
arctan(
√ 5 2
arctan(
√ 5 2
π
Problem 7 (10 pts) Find the integral
2 + x
2 − x
dx.
SOLUTION
2 + x
2 − x
dx =
2 + x √ 4 − x^2
dx =
4 − x^2
dx +
x √ 4 − x^2
dx
Using substitution, x = 2 sin u and dx = 2 cos u du.
∫ 2 √ 4 − x^2
dx =
2 cos u
2 cos u du =
2 du = 2u = 2 arcsin(
x
x √ 4 − x^2
dx = −
4 − x^2
2 + x
2 − x
dx =
4 − x^2
dx +
x √ 4 − x^2
dx
= 2 arcsin(
x
4 − x^2 + C
