HwSln 5

HW 5 (Integration by parts – Due Sep 26, Wed
1. Find indefinite integrals
Z
(a)
xe2x dx
dv = e2x dx
1
• du = dx, v = e2x
Z2
1 2x
1 2x
1
1
• Soln= xe −
e dx = xe2x − e2x + C
2
2
2
4
• u = x,
Z
(b)
(x + 1)e2x dx
dv = e2x dx
1
• du = dx, , v = e2x
2 Z
1 2x
1
e dx
• Soln= (x + 1)e2x −
2
2
1
1
• = (x + 1)e2x − e2x + C
2
4
Z
√
√
x ln xdx
(c)
• u = (x + 1),
√
1
• u = ln x = ln x, dv = x1/2 dx
2
1 −1
2
• du = x , v = x3/2
2
3 Z
Z
1
1
1
1
3/2
−1 3/2
3/2
x x dx = ln x · x −
x1/2 dx
• Soln= ln x · x −
3
3
3
3
1
2
• = ln x · x3/2 − x3/2 + C
3
9
Z
ln x
(d)
dx
x2
• u = ln x, dv = x−2 dx
• du = x−1 , v = −x−1
Z
Z
−1
−1
−1
−1
• Soln=− ln x · x − x (−x )dx = − ln x · x + x−2 dx
Z
(e)
• =− ln x · x−1 − x−1 + C
x
√
dx
2x + 3
1
dv = (2x + 3)−1/2 dx
1 (2x + 3)1/2
• du = dx, v =
= (2x + 3)1/2
2
1/2
Z
1 (2x + 3)3/2
• Soln=x(2x + 3)1/2 − (2x + 3)1/2 dx = x(2x + 3)1/2 −
2
3/2
3/2
(2x + 3)
+C
• = x(2x + 3)1/2 −
3
• u = x,
Z
(f)
3x2 (ln x)2 dx
•
•
•
•
use IBP two times.
1st time
u = (ln x)2 , dv = 3x2 dx
du = 2 ln xdx, v =Z x3
• Soln=(ln x)2 x3 − 2
ln x · x3 dx
Z
• 2nd time calculating ln x · x3 dx
dv = x3 dx
x4
−1
du = x , v =
4
Z
Z
1
1
3
4
ln x · x dx = ln x · x −
x−1 x4 dx
4
4
1
1
= ln x · x4 − x4
4
16 1 4
1
4
2 3
ln x · x − x
Soln= (ln x) x − 2
4
16
• u = ln x,
•
•
•
•
Z
(g)
√
e−
x
dx. Hint: first, make the substitution t =
√
1
• t = x, dt = √ dx,
2 x
Z
Z
√
•
e− x dx = e−t 2tdt
√
dx = 2 xdt = 2tdt
• u = 2t, dv = e−t dt
−t
• du = 2dt, v = −e
Z
• Soln=−2te−t − 2
Z
(h)
√
x then integration by parts.
√
√
√
(−e−t )dt = −2te−t − 2e−t + C = −2 xe− x − 2e− x + C
x2 sin(2x)dx
2
• Use IBP two times
• 1st time:
• u = x2 , dv = sin(2x)dx
1
• du = 2xdx, v = − cos(2x)
2 Z
Z
1 2
1
1 2
• Soln=− x cos(2x) −
2x[− cos(2x)]dx = − x cos(2x) + x cos(2x)dx
2
2
Z2
• 2nd time: calculate
x cos(2x)dx
• u = x,
•
•
•
•
dv = cos(2x)dx
1
du = dx, v = sin(2x)
2
Z
Z
1
1
x cos(2x)dx = x sin(2x) −
sin(2x)dx
2
2
1
1
= x sin(2x) + cos(2x)
2
4
1 2
1
1
Soln= − x cos(2x) + x sin(2x) + cos(2x) + C
2
2
4
2. The velocity of a dragster t sec after leaving the starting line is
100te−0.2t ft/sec
What is the distance covered by the dragster int he first 10 sec of its run?
Soln:
Z
10
D=
100te−0.2t dt
0
dv = e−0.2t dt
1 −0.2t
du = 100dt, v =
e
= −5e−0.2t
−0.2
Z
u = 100t,
F (t) =
−0.2t
100te
dt = −500te
−0.2t
F (t) = −500te−0.2t − 2500e−0.2t
F (10) = −1015, F (0) = −2500
D = F (10) − F (0) = 1485 fts
3
+ 500
Z
e−0.2t dt