### Solution

```Problem Set 9
Due 2:00PM Sep. 14
1. Evaluate the following integrals
R √2
• 1 xdx
3
R√
7
• 0 x2
R 1/2 2
• 0 t dt
R 2a
• a xdx
√
R 3b
• 0 x2 dx
R1
• 3 7dx
R2
• 0 5xdx
R2
• 0 (2t − 3)dt
R1
z
• 2 1 + 2 dz
R2 2
• 1 3u du
R2
• 0 (3x2 + x − 5)dx
2. Find the indefinite integral of each of the following functions.
• 4x6 − x3 ⇒ 74 x7 − 41 x4 + c
• e(ex2+6x) (x + 1)
• 12x2 − 6x1/2 + 3x−1/2 − x−1
3
1
⇒ 4x3 − 4x 2 + 6x 2 − lnx + c
• (x2 + 2x + 4)1/2 (x + 1)
3
⇒ 13 (x2 + 2x + 4) 2 + c
• 6e7x ⇒ 67 e7x + c
•
3. Use integration by parts to integrate
•
R
xlnxdx ⇒ 21 x2 lnx − 41 x2
1
3x1/2 +x−1/2
x3/2 +x1/2
•
R
x2 e2x dx ⇒ e2x ( 12 x2 − 12 x + 41 )
4. Find the indicated integral
•
R
1
dx
x7
•
⇒ − 6x16 + c
R
(3x2 −
x3 −
•
R
1
x−a dx
⇒ log|x − a| + c
•
R
√
5x + 2)dx ⇒
√
2 5 32
3 x
(7x3/5 −
+ 2x + c
√
x7 )dx ⇒
5. Integration by parts
• Find
R3
2
ln xdx
Z
3
Z
3
ln xdx =
ln x1dx
Z
= x ln x|32 −
2
2
= (x ln x −
3
1dx
2
x)|32
= 3 ln 3 − 2 ln 2 − 1
• Find
R
x6 ln xdx
Z
• Find
R
x6 ln xdx =
Z
(ln x)x6 dx
Z
1
1 71
= ln x( x7 ) −
x dx
7
7 x
Z
1 7
1
=
x ln x −
x6 dx
7
7
1 7
1
=
x ln x − x7 + C
7
49
x2 e3x dx.
2
35 58
8 x
9
− 92 x 2 + c
Let u = x2 and dv = e3x dx.R Thus du = 2xdx and v = 31 e3x . We then integrate by
parts and obtain 13 x2 e3x − 32 xe3x dx. Now integrate by parts again with u = x and
dv = e3x dx. Thus:
Z
Z
1 2 3x 2
2 3x
xe3x dx
x e −
x e dx =
3
3
Z
1 2 3x 2 1 3x
1 3x
=
x e − {x e −
e dx} Now have e and x seperate!
3
3 3
3
1 2 3x 2 3x
2
=
x e − xe + e3x + C
3
9
27
• Show that
• Find
R
R
xn ex dx = xn ex − n
R
xn−1 ex dx
ln x
dx.
x5
Let u = ln x, dv =
= x−5 dx. This means that du = x1 dx and v =
Z
Z
ln x
−1
−1 1
dx = ln x( 4 ) − ( 4 ) dx
5
x
4x
4x x
−4
1x
−1
+C
= ln x( 4 ) +
4x
4 −4
−1
x−4
ln x( 4 ) −
+C
4x
16
1
dx
x5
3
−1
.
4x4
Thus
```