MATH 19B Midterm: MATH19B Midterm 2014 Fall Solution
Midterm Solutions for Math 19B, Verison 1
1. a. R1
0x10dx = [x11
11 ]1
0=1
11 .
b. Let u=x3. Then du = 3x2dx. We now have that
Zx2sin(x3)dx =Z1
3sin(u)du =−
1
3cos(u) + C=−
1
3cos(x3) + C
c. Let u= tan x. Then du = (sec2x)dx. Now,
Zπ
4
0
(sec2x)etan xdx =Z1
0
eudu =e−1
d. Let u= ln x. Then du =1
xdx.
Then
Z1
x(ln x)2dx =Zu−2du =−
1
u+C=−
1
ln x+C.
2. a. V=πR1
0(ex)2dx =πR1
0e2xdx
b. V=πR1
0e2xdx = [πe2x
2]1
0=π
2(e2
−1).
c. 2πR1
0xexdx
d. 2πR1
0xexdx = 2π[xex
−ex]1
0= 2π
3. a. R10
5(2f(x)−g(x))dx = 2 R10
5f(x)dx −
1
7R10
5g(x)dx = 2(10) −3 = 17.
b. R5
10 f(x)dx =−R10
5f(x)dx =−10.
4. a. ln(12 + cos x).
b. cos xp1 + (sin x)3
5. a. A(x) = 2(1 −x2). To see this, notice that the triangle has base equal
to 2√1−x2. The height is equal to the base and so the area of the triangle
equals 1
2·(2√1−x2)2= 2(1 −x2).
b. V=R1
−1A(x)dx =R1
−12(1 −x2)dx.
c. V=R1
−12(1 −x2)dx = [2x−
2x3
3]1
−1=8
3.
6. Rxsin xdx =−xcos x+ sin x+C. This problem can be solved using
integration by parts with u=xand dv = sin xdx.
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MATH 19B Full Course Notes
Verified Note
17 documents
Document Summary
11: a. (cid:82) 1 (cid:90) (cid:90) 1 (cid:90) . We now have that x2 sin(x3)dx = sin(u)du = 1. 3 cos(x3) + c: let u = tan x. 0 eudu = e 1 u 2du = 1 u. 5 g(x)dx = 2(10) 3 = 17. 0 e2xdx = [ e2x: let u = ln x. 0 xexdx = 2 [xex ex]1 (cid:90) (cid:90) 0 (ex)2dx = (cid:82) 1: a. v = (cid:82) 1, v = (cid:82) 1, 2 (cid:82) 1, 2 (cid:82) 1. 5 (2f (x) g(x))dx = 2(cid:82) 10: a. (cid:82) 10. 10 f (x)dx = (cid:82) 10: (cid:82) 5, cos x(cid:112)1 + (sin x)3. 1 a(x)dx =(cid:82) 1: a. ln(12 + cos x). 1 x2)2 = 2(1 x2): a. To see this, notice that the triangle has base equal. The height is equal to the base and so the area of the triangle to 2.
