Download Exam II Solutions 1. (a) lim sin x → 0 0 − sin(x)e cos x = − sin(0) cos

Exam II Solutions
1. (a)
ecos(x) − e
0
→
x→0
sin x
0
lim
− sin(x)ecos(x)
− sin(0)
=
=0
x→0
cos x
cos(0)
=L’Hop lim
1. (b)
lim (1 − x)(1−x) → 00
x→1−
lim (1 − x) ln(1 − x) = lim
x→1−
x→1−
=L0 Hop lim−
x→1
−∞
ln(1 − x)
→
1/(1 − x)
∞
1/(1 − x)
= lim− −(1 − x) = 0
−1/(1 − x)2
x→1
lim (1 − x)(1−x) = e0 = 1
x→1−
2. (a) Let u = (1 − 5x100 ) so that du = −500x99 dx. The answer will then be given
by
Z
x99
1
dx = −
(1 − 5x100 )2
500
Z
u−2 du =
1 1
1
1
+C =
+ C.
500 u
500 (1 − 5x100 )
(c) Let u = tan t so that du = sec2 tdt giving
Z
π/4
tan t
e
2
Z
sec tdt =
0
0
1
1
e du = e = e − 1.
u
u
0
(d) The average value is given by
1
(π/4 − 0)
Z
π/4
etan t sec2 tdt =
0
1
1
4(e − 1)
(e − 1) =
.
π/4
π
2
3. This is a straightforward use of FTC along with the chain rule
Z ex
g(x) =
sin(ln t)dt
10000.5
g 0 (x) = sin(ln(ex )) · (ex )0 = sin(x) · ex
4. The area is given by the expression
3
Z 3
Z 3
2x3
2
2
2
(9 − x ) − (x − 9)dx =
18 − 2x dx = 18x −
= 72.
3 −3
−3
−3
5. (a) The function is increasing on (0, +∞) and decreasing on (−∞, 0).
(b) There is one critical point at x = 0, and it is a local minimum. Because
this is the only critical point, it is in fact a global minimum.
Possible inflection
√
points occur when f 00 = 0 which happens when x = ±1/ 3. The second derivative
switches sign at both of these points, and so they are in fact points of inflection.
(c) The function will never divide by zero, and so will never have a vertical asymptote. The line y = 1 is a horizontal asymptote because
x2
= 1.
x→±∞ x2 + 1
lim
(d) Your sketch should reflect the above information accurately, noting that, at
x = 0, f (x) = f (0) = 0.
6. The limit can be recognized as an integral
1
1999
Z 1
n
X
i
1
lim
2000 ·
=
2000x1999 dx = x2000 = 1.
n→∞
n
n
0
i=1
0
7. Note: there are TWO mistakes made in this computation:
1
Z 1
1 2
1
dx = − 3 = −
2
x
3x
3
−1
−1
is incorrect, first and most importantly, because the fundamental theorem of calculus was applied to compute the definite integral. The function 1/x2 is NOT
continuous on the interval of integration, and so we do not know that FTC can be
applied!
The second, less offensive mistake is that the fundamental theorem of calculus was
INCORRECTLY applied! Specifically, as many of you noted, the function −1/3x3
is NOT an anti-derivative for 1/x2 .
3
Even though I only intended there to be one mistake (the first one), either observation was worth full credit.