Math 2143 - Brief Calculus with Applications
Test #2 - 2013.03.26
Solutions
Compute the following derivatives.
1.
d x2 +3x−9
4
dx
2
d x2 +3x−9
4
= ln(4) 4x +3x−9 (2x + 3)
dx
2.
d x2
4 + 43x + 4−4
dx
2
d x2
4 + 43x + 4−4 = ln(4) 4x 2x + ln(4) 43x 3
dx
3.
2x+1
d 3x−7
e
dx
2x+1
2x+1 2(3x − 7) − (2x + 1)3
d 3x−7
e
= e 3x−7
dx
(3x − 7)2
4.
d e2x+1
dx e3x−7
d e2x+1
d 2x+1−(3x−7)
e
=
3x−7
dx e
dx
d −x+8
=
e
dx
= −e−x+8
5.
√
d
ln (3x + 4) 3x + 1
dx
√
d
d
ln (3x + 4) 3x + 1 =
dx
dx
1
ln(3x + 4) + ln(3x + 1)
2
3
1 3
=
+
3x + 4 2 3x + 1
6.
√
d
ln (3x + 4) + 3x + 1
dx
3
3 + 2√3x+1
√
d
√
ln (3x + 4) + 3x + 1 =
dx
(3x + 4) + 3x + 1
7.
√
d
ln(3x + 4) + ln 3x + 1
dx
The answer here is the same as that of problem 5.
2
√
d 2x+1
e
log2 ( 3x + 1)
dx
8.
√
d 2x+1
e
log2 ( 3x + 1) =
dx
√
√
d 2x+1
d
e
log2 ( 3x + 1) + e2x+1
log2 ( 3x + 1)
dx
dx
√
1
1
3
= 2 e2x+1 log2 ( 3x + 1) + e2x+1
ln(2) 2 3x + 1
Compute the following itegrals.
Z
t2 (t3 +1)12 dt
9.
Here, we let u = t3 + 1, thus du = 3t2 dt, or t2 dt = 13 du.
Z
Z
1
2 3
12
u12 du
t (t + 1) dt =
3
1 13
u +C
=
39
1 3
=
(t + 1)13 + C
39
Z
(ln(x))3
10.
dx
x
Here is another substitution, we let u = ln(x), so du = x1 dx:
Z
Z
(ln(x))3
dx = u3 du
x
1
= u4 + C
4
1
= (ln(x))4 + C
4
Z
11.
z 2 (ln(z))2 dz
Integration by parts here, du = z 2 dz, and v = (ln(z))2 , which gives u = 13 z 3 and dv = 2 ln(z) z1 dz. So we have
Z
Z
1
2
z 2 (ln(z))2 dz = z 3 (ln(z))2 −
z 2 ln(z)dz
3
3
We do not know the integral on the right, so we do another integration by parts, with du = z 2 dz, and v = ln(z).
Thus, u = 31 z 3 and dv = z1 dz. Our integral now looks like:
Z
Z
1
2
z 2 (ln(z))2 dz = z 3 (ln(z))2 −
z 2 ln(z)dz
3
3
Z
1 3
2 1 3
1
2
2
= z (ln(z)) −
z ln(z) −
z dz
3
3 3
3
1
2 1 3
1
= z 3 (ln(z))2 −
z ln(z) − z 3 + C
3
3 3
9
1 3
2 3
2 3
2
= z (ln(z)) − z ln(z) + z + C
3
9
27
Z
1
12.
−1
p
4 − 4w2 dw
3
First, we write this as
Z
1
p
4 − 4w2 dw = 2
1
Z
p
1 − w2 dw
−1
−1
Note that the integral on the right is the area of the upper 1/2 circle of radius one, thus
Z 1p
Z 1p
4 − 4w2 dw = 2
1 − w2 dw
−1
−1
=π
Z
8
13.
√
x x + 1dx
0
Using the substitution u = x + 1, we have du = dx and x = u − 1:
Z 8
Z 8
√
√
(u − 1) udu
x x + 1dx =
0
0
Z
=
8
u3/2 − u du
0
8
2 5/2 1 2 u − u 5
2 0
2 5/2 1 2
= 8 − 8
5
2
=
Z
3
14.
3
y − y dy
−3
So first we need to determine how to define y 3 − y = |y(y − 1)(y + 1)|:
 3
y − y,
y≥1



3
−(y 3 − y), 0 ≤ y < 1
y − y =
y 3 − y,
−1 ≤ y < 0



3
−(y − y), y < −1
Next up, since we have an odd function, when we take an absolute value, we have an even function, thus:
Z 3
Z 3
3
3
y − y dy = 2
y − y dy
−3
0
We now use piecewise definitions given above:
Z 3
Z
3
y − y dy = 2
−3
3
3
y − y dy
0
Z
=2
=
65
2
3
Z
−(y − y)dy +
3
3
y − y dy
1
3 !
1 4 1 2 1 4 1 2 − y + y + y − y 4
2 0
4
2 1
0
=2
1
1