Math 234
April 19
I. Use integration by parts to find the given integral
1.
Re
2.
R
3.
R1
4.
R
x(ln(x))2 dx
5.
R
ln(x)
dx
x4
6.
R1
7.
R1
8.
R1
9.
R2
10.
R
1
x ln(x)dx
x5/2 ln(x)dx
0
0
0
0
1
x3 ex dx
x(e−5x + ex/2 )dx
√ x dx
x+1
2
√x
dx
x+2
x(e−5x + ln(x))dx
5
√x
dx
x2 +1
II.Describe the domain of the given function
1. f (x, y) =
x+y
x−y
2. f (x, y) =
x+y
x+ x−y
1
3. f (x, y) =
x+y
x2 +y 2
4. f (x, y) =
p
x2 − y 2
5. f (x, y) = ln(x2 − y)
x+y
6. f (x, y) = √e 2
x −y 2
III. Sketch the indicated level curve f (x, y) = C for each choice of constant
C.
1. f (x, y) = x2 + y; C = 0, C = 4, C = 9
2. f (x, y) = x/y; C = −2; C = 2
3. f (x, y) = yex ; C = 0; C = 1
4. f (x, y) = xey ; C = 0; C = 1
IV. Find all first-order partial derivatives
1. f (x, y) = x2 + y
2. f (x, y) = x/y
3. f (x, y) =
x+y
x−y
4. f (x, y) =
p
x2 − y 2
x+y
5. f (x, y) = √e 2
x −y 2
2
I. Use integration by parts to find the given integral
Re
1.
x ln(x)dx
1
Answer:
u = ln(x)
du = x1 dx
v = x2 /2
dv = xdx
R e x2
Re
x2 ln(x) e
|
−
1xdx
x
ln(x)dx
=
1
2
1 2
1
Re
e2 ln(e)
12 ln(1)
= 2 − 2 − 12 1 xdx
2
2
= e2 − 21 x2 |e1
2
= e2 − 41 (e2 − 1)
2
= e4 + 14
2.
R
x5/2 ln(x)dx
Answer:
u = ln(x)
du = x1 dx
7/2
dv = x5/2 dx
v = x7/2
R 5/2
R 7/2
7/2
x ln(x)dx = 2x 7ln(x) − 2x7 x1 dx
R
7/2
= 2x 7ln(x) − 27 x5/2 dx
7/2
7/2
= 2x 7ln(x) − 27 x7/2 + C
=
3.
2x7/2 ln(x)
7
−
4 7/2
x
49
+C
R1
x3 ex dx
Answer:
u = x3
du = 3x2 dx
v = ex
dv = ex dx
R 3 x
R
x e dx = x3 ex − R3x2 ex dx
u = x2
= x3 ex − 3 x2 ex dxR
v = ex
= x3 ex − 3(x2 ex − R 2xex dx
= x3 ex − 3x2 ex + 6 xex dxR
u =x
= x3 ex − 3x2 ex + 6(xex − ex dx)
v = ex
= x3 ex − 3x2 ex + 6xex − 6ex + C
R1 3 x
x e dx = x3 ex − 3x2 ex + 6xex − 6ex |10
0
= e1 − 3e1 + 6e1 − 6e1 − (0 − 0 + 0 − 6e0 )
= −2e1 + 6
0
3
du = 2xdx
dv = ex dx
du = dx
dv = ex dx
4.
5.
R
x(ln(x))2 dx
Answer:
u = (ln(x))2
du = 2 ln(x)(1/x)dx
2
v
= x /2
dv = xdx
R
R 2
2
2
x(ln(x))2 dx = x (ln(x))
− x2 2 ln(x)(1/x)dx
2
R
2
2
− x ln(x)dx
= x (ln(x))
2
R 2
2
2
2
= x (ln(x))
− ( x ln(x)
− x2 x1 dx)
2
2
R
2
2
x2 ln(x)
1
−
+
xdx
= x (ln(x))
2
2
2
x2 (ln(x))2
x2 ln(x)
1 x2
=
− 2 + 2 2 +C
2
2
2
x2 (ln(x))2
=
− x ln(x)
+ x4 + C
2
2
u = ln(x)
v = x2 /2
du = x1 dx
dv = xdx
ln(x)
dx
x4
R
Answer:
u = ln(x)
du = (1/x)dx
v = x−3 /−3
dv = x−4 dx
R −1 1
R ln(x)
ln(x)
dx = −3x
− 3x
3
3 x dx
x4
R
− ln(x)
= 3x3 + (1/3) x−4 dx
−3
ln(x)
= −3x
+ 13 x−3 + C
3
ln(x)
− 9x13 + C
= −3x
3
6.
R1
7.
R1
x(e−5x + ex/2 )dx
Answer:
u
=x
du = dx
−5x
x/2
v = (−1/5)e
+ 2e
dv = e−5x + ex/2 dx
R1
R1
x(e−5x + ex/2 dx = x((−1/5)e−5x + 2ex/2 )|10 − 0 (−1/5)e−5x + 2ex/2 dx
0
−5
= −e5 + 2e1/2 − (1/25)e−5x + 4ex/2 |10
−5
= −e5 + 2e1/2 − ((1/25)e−5 + 4e1/2 − (1/25 + 4))
−5
= −6e
+ −2e1/2 + 4.04
25
0
0
√ x dx
x+1
Answer:
u
=x
v = 2(x + 1)1/2
du = dx
dv = (x + 1)−1/2 dx
4
R1
√ x dx
0
x+1
8.
R1
0
R1
= 2x(x + 1)1/2 |10 − 0 2(x + 1)1/2 dx
R1
= 2(1 + 1)1/2 − 0 − 2 0 (x + 1)1/2 dx
= 2(2)1/2 − 2 23 (x + 1)3/2 |10
= 2(2)1/2 − 43 (23/2 − 1)
2
√x
dx
x+2
Answer:
u
= x2
du = 2xdx
1/2
v = 2(x + 2)
dv = (x + 2)−1/2 dx
R 1 x2
R1
√
dx = 2x2 (x + 2)1/2 |10 − 0 2x2(x + 2)1/2 dx
0
x+1
R1
= 2(1 + 2)1/2 − 0 − 4 0 x(x + 2)1/2 dx
R1
3/2
1
= 2(3)1/2 − 4( 2x(x+2)
−
(2/3)(x + 2)3/2 dx)
|
0
3
0R
3/2
3/2
1
= 2(3)3 − 4( 2(1+2)
− 0 − 23 0 (x + 2)3/2 dx)
3
3/2
3/2
= 2(3)3 − 43 ( 2(3)3 − 23 25 (x + 2)5/2 |10 )
= 32 (3)3/2 − 89 (3)5/2 − 16
((3)5/2 − 1)
45
= 23 (3)3/2 − 89 (3)5/2 − 16
(3)5/2 + 16
)
45
45
9.
10.
R2
x(e−5x + ln(x))dx
Answer:
R2
R2
R2
x(e−5x + ln(x))dx = 1 x(e−5x dx + 1 x ln(x)dx
1
R
R2
−5x
From pervious problems we have: x(e−5x dx+ 1 x ln(x)dx = −xe5 −
2
2
− x4 + C
e−5x + x ln(x)
2
R2
2
−5x
2
x(e−5x + ln(x))dx = ( −xe5 − e−5x + x ln(x)
− x4 )|21
2
1
−10
−5
= −2e5 − e−10 + 4 ln(4)
− 44 ) − ( −e5 − e−5 − 14 )
2
−5
−10
= −7e5 + 2 ln(4) − 1 + 6e5 + 14
1
R
5
√x
dx
x2 +1
Answer:
u
= x4
v = (1/2)(x2 + 1)−1/2
u
= x2
v = (3/2)(x2 + 1)3/2
du = 4x3 dx
dv = x(x2 + 1)−1/2 dx
du = 2xdx
dv = 2x(x2 + 1)1/2 dx
5
R
5
√x
dx
x2 +1
=
=
=
=
√
R
x4 (x2 +1)1/2
− 2x3 x2 + 1dx
2
R
x4 (x2 +1)1/2
− ( 32 x2 (x2 + 1)3/2 − 2x 23 (x2 + 1)3/2 dx
2
x4 (x2 +1)1/2
− 32 x2 (x2 + 1)3/2 + 23 25 (x2 + 1)5/2 + C
2
1 4 2
4
x (x + 1)1/2 − 23 x2 (x2 + 1)3/2 + 15
(x2 + 1)5/2 + C
2
II.Describe the domain of the given function
1. f (x, y) = x+y
x−y
Answer:
x − y 6= 0
All (x, y) pairs where x and y are real numbers such that y 6= x.
{(x, y) ∈ R2 |y 6= x}
2. f (x, y) = x2x+y
+x−y
Answer:
x2 + x − y 6= 0
All (x, y) pairs where x and y are real numbers such that y 6= x2 − x.
{(x, y) ∈ R2 |y 6= x2 − x}
3. f (x, y) = xx+y
2 +y 2
Answer:
x2 + y 2 6= 0
All (x, y) pairs where x and y are real numbers such that |y| =
6 |x|.
2
{(x, y) ∈ R ||y| =
6 |x|}
p
4. f (x, y) = x2 − y 2
Answer:
x2 − y 2 ≥ 0
All (x, y) pairs where x and y are real numbers such that |x| ≥ |y|.
{(x, y) ∈ R2 ||x| ≥ |y|}
6
5. f (x, y) = ln(x2 − y)
Answer:
x2 − y ≥ 0
All (x, y) pairs where x and y are real numbers such that x2 ≥ y.
{(x, y) ∈ R2 |x2 ≥ y}
x+y
6. f (x, y) = √e 2
x −y 2
Answer:
x2 − y 2 > 0
All (x, y) pairs where x and y are real numbers such that |x| > |y|.
{(x, y) ∈ R2 ||x| > |y|}
III. Sketch the indicated level curve f (x, y) = C for each choice of constant
C.
1. f (x, y) = x2 + y; C = 0, C = 4, C = 9
Answer:
9
8
6
4
4
2
0
0
-2
-4
-3
-2
-1
0
1
2
7
3
2. f (x, y) = x/y; C = −2; C = 2
Answer:
2
1
0
-1
-2
-3
-2
-1
0
1
2
3
2
3
3. f (x, y) = yex ; C = 0; C = 1
Answer:
2
1
1
0
0
-1
-2
-3
-2
-1
0
1
4. f (x, y) = xey ; C = 0; C = 1
8
Answer:
3
2
1
0
-1
1
0
-2
-1
0
1
2
3
IV. Find all first-order partial derivatives
1. f (x, y) = x2 + y
Answer:
fx (x, y) = 2x
fy (x, y) = 1
2. f (x, y) = x/y
Answer:
fx (x, y) = y1
fy (x, y) = −x
y2
3. f (x, y) = x+y
x−y
Answer:
=
fx (x, y) = 1(x−y)−(x+y)(1)
(x−y)2
fy (x, y) =
−2y
(x−y)2
1(x−y)−(x+y)(−1)
2x
= (x−y)
2
(x−y)2
9
4
p
4. f (x, y) = x2 − y 2
Answer:
fx (x, y) = (1/2)(x2 − y 2 )−1/2 (2x)
fy (x, y) = (1/2)(x2 − y 2 )−1/2 (−2y)
x+y
5. f (x, y) = √e 2
x −y 2
Answer:
fx (x, y) =
√
ex+y (1)(
x2 −y 2 )−ex+y (1/2)(x2 −y 2 )−1/2 (2x)
√
( x2 −y 2 )2
p
= x2 −y2 ( x2 − y 2 ) − x(x2 − y 2 )−1/2 )
√
ex+y (1)( x2 −y 2 )−ex+y (1/2)(x2 −y 2 )−1/2 (−2y)
√
fy (x, y) =
( x2 −y 2 )2
p
x+y
= xe2 −y2 ( x2 − y 2 ) + y(x2 − y 2 )−1/2 )
ex+y
10