MATH 101 SAMPLE MIDTERM 2 – ANSWERS
1. (a) Integrate by parts with f = x, f 0 = 1, g 0 = e−2x , g = −e−2x /2:
Z
1
1
xe−2x dx = (− x − )e−2x + C.
2
4
Z
p
1
9 − x2 dx = − (9 − x2 )3/2 + C (can substitute u = 9 − x2 , or else substitite
3
x = sin θ and work from there).
Z
ex dx
(c)
= ln |ex − 2| − ln |ex − 1| + C, by substituting u = ex and then using
2x
x
e − 3e + 2
partial fractions.
(b)
x
Z
1/2
2. (a)
Z
(b)
0
∞
dx
diverges to −∞.
x ln x
(x3 + x)−1/2 dx is convergent. (These two examples were done in class.)
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3. We want to evaluate numerically the integral
Z
2
1
sin x
dx
x
using the Midpoint Approximation.
(a)
M5 = 0.2
sin 1.1
1.1
+
sin 1.3 sin 1.5 sin 1.7 sin 1.9 +
+
+
.
1.3
1.5
1.7
1.9
(b) The error estimate for the Midpoint Rule is K(b − a)2 /24n2 , where K = max |f 00 (x)|
on [a, b]. If f (x) = (sin x)/x, then
f 00 (x) = −
sin x + cos x sin x
+ 2
x
x
so we can take K = 3. For the error to be less than 10−5 , we need
3 · 12
≤ 10−5 ,
24n2
√
i.e. n ≥ 50 5. (If this was a homework problem, you would be expected to compute this
using a calculator, but on a midterm you don’t have to do this.)
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