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Answers to Review Sheet for Final Exam

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(d) d<br />

dy<br />

<br />

y + 1<br />

y + 7<br />

<br />

y + 7<br />

Solution: 3<br />

y + 1<br />

1<br />

(y + 7) 2<br />

41. Calculate the following limits (make sure <strong>to</strong> justify your answers).<br />

ln x<br />

(a) lim √<br />

x→1 x + 1<br />

Solution: Just plug in x = 1 <strong>to</strong> get 0/2 = 0.<br />

(b) lim<br />

x→∞ x 2 e −x<br />

Solution: Use L’Hopital’s rule <strong>to</strong> get<br />

lim<br />

x→∞<br />

x2 = lim<br />

ex x→∞<br />

2x<br />

= lim<br />

ex x→∞<br />

2<br />

= 0.<br />

ex (c) lim x ln x<br />

x→0 +<br />

Solution: It’s of the <strong>for</strong>m 0 · ∞, so we rewrite as a fraction:<br />

42. Evaluate the following integrals.<br />

(a)<br />

2<br />

0<br />

4 − y 2 dy<br />

lim x ln x = lim<br />

x→0 + x→0 +<br />

ln x<br />

1/x<br />

= lim<br />

x→0 +<br />

1/x<br />

= lim<br />

−1/x2 x→0 +(−x)<br />

= 0.<br />

Solution: You’re supposed <strong>to</strong> recognize this as the area of a quarter-circle of radius 2,<br />

which is 1<br />

4 (π · 22 ) = π.<br />

√<br />

t + t + 1<br />

(b)<br />

t2 dt<br />

√<br />

t + t + 1<br />

Solution:<br />

t2 <br />

t−1 −3/2 −2<br />

dt = + t + t dt = ln |t| − 2t −1/2 − t −1 + C<br />

2 θ<br />

(c) 2 + 1 dθ<br />

1<br />

Solution:<br />

θ 2<br />

<br />

θ=2 2<br />

1<br />

2 2<br />

It’s + θ = + 2 − + 1 =<br />

ln 2 θ=1 ln 2 ln 2 2<br />

+ 1.<br />

ln 2<br />

43. Calculate<br />

2<br />

1<br />

5<br />

1<br />

[3f(x) + 4g(x)] dx, given that<br />

f(x) dx = −1,<br />

5<br />

2<br />

f(x) dx = 1,<br />

10<br />

5<br />

−1<br />

g(x) dx = −2, and<br />

1<br />

−1<br />

g(x) dx = 3.

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