Due: 04/07/2016 in class
Instructor: Janina Letz
MATH 1210, Spring 2016
Lab #11
Name:
Question:
1
2
3
4
5
Total
Points:
8
8
20
4
16
56
Score:
Instructions: Please show all of your work as partial credit will be given where appropriate, and there may
be no credit given for problems where there is no work shown. All answers should be completely simplified,
unless otherwise stated. No calculators or electronics of any kind are allowed.
1. (8 points) Let f be continuous on [a, b] and integrable there. Show that
Z
Z
b
b
|f (x)|dx
f (x)dx ≤
a
a
(Hint: Find an upper and a lower bound for f (x) and use the comparison property)
Solution: Since −|f (x)| ≤ f (x) ≤ |f (x)| for all x ∈ [a, b], then by Theorem B we have
b
Z
−
b
Z
|f (x)|dx ≤
a
Z
f (x)dx ≤
a
so
b
|f (x)|dx
a
Z
Z
b
b
f (x)dx ≤
|f (x)|dx
a
a
2. (8 points) Suppose that f 0 is integrable and |f 0 (x)| ≤ M for all x. Prove that |f (x)| ≤ |f (a)| + M |x − a|
(Hint: Integrate |f 0 (x)| from a to x and use the comparison property)
Solution: Assume x ≥ a. Since |f 0 (x)| ≤ M , we have that
Z x
Z x
|f 0 (t)|dt ≤
M dt
a
With problem 1 this gives
Z x
Z
0
|
f (t)dt| ≤
a
x
0
a
x
Z
|f (t)|dt ≤
a
M dt
|f (x) − f (a)| ≤ M (x − a)
a
The triangle inequality gives
|f (x)| − |f (a)| ≤ |f (x) − f (a)| ≤ M (x − a)
|f (x)| ≤ |f (a)| + M (x − a)
If x < a, then the prove is the same with x and a switched, but for the last step, there use
|f (x)| − |f (a)| ≤ |f (a) − f (x)|. So the resulting inequality for this case is
|f (x)| ≤ |f (a)| + M (a − x)
So for both cases the claim is true.
3. (First fundamental theorem) Find the derivatives of the given functions without evaluating the
integrals.
1
MATH 1210, Spring 2016
Lab #11
Z
b
t3 dt where a and b are real numbers
(a) (4 points) f (x) =
a
Solution:
f 0 (x) = 0
Z
x
(b) (4 points) f (x) =
p
u2 + 2du
−x
Solution: Let F (x) =
Rx√
0
u2 + 2du. Then
p
f 0 (x) = F 0 (x) + F 0 (x) = 2 x2 + 2
f (x) = F (x) − F (−x)
Z
x+2
(4t + 1)dt
(c) (4 points) f (x) =
x
Solution: Let F (x) =
Rx
0
(4t + 1)dt. Then
f 0 (x) = F 0 (x + 2) − F 0 (x) = 4(x + 2) + 1 − (4x + 1) = 8
f (x) = F (x + 2) − F (x)
Z
x+1
(d) (4 points) f (x) =
−x2
t2
dt
1 + t2
Solution: Let F (x) =
Rx
t2
dt.
0 1+t2
f 0 (x) = F 0 (x + 1) + 2xF 0 (−x2 ) =
f (x) = F (x + 1) − F (−x2 )
Z
Then
(x + 1)2
2x5
+
1 + (x + 1)2
1 + x4
tan(x)
t4 dt
(e) (4 points) f (x) =
sec(x)
Solution: Let F (x) =
Rx
0
t4 dt. Then
f (x) = F (tan(x)) − F (sec(x))
f 0 (x) = sec(x)2 F 0 (tan(x)) − sec(x) tan(x)F 0 (sec(x))
= sec(x)2 tan(x)4 − sec(x)5 tan(x)
4. (4 points) (Integral graphically) The graph shows a function, whose third derivative is continuous.
Decide wether the following integrals are positive or negative.
4
3
2
1
0
2
0
1
2
3
4
Due: 04/07/2016 in class
Instructor: Janina Letz
i.
R3
ii.
R3
0
0
MATH 1210, Spring 2016
Lab #11
Name:
f (x)dx
iii.
√
positive
negative
R3
f 0 (x)dx
iv.
√
positive
negative
R3
0
0
f 00 (x)dx
positive
√
negative
f 000 (x)dx
√
positive
negative
5. (Substitution Rule) Solve the following integrals using substitution.
Z
p
(a) (4 points)
−2x 9 − x2 dx
Solution: Let u = 9 − x2 , du = −2 x dx. Then
Z
Z
p
√
3
2
2 3
−2x 9 − x2 dx =
u du = u 2 + C = (9 − x2 ) 2 + C
3
3
Z
4
x(4x2 + 3)3 dx
(b) (4 points)
−2
Solution: Let u = 4 x2 + 3, du = 8 x dx. Then
Z 4
Z
1 67 3
1 u4
2
3
x(4x + 3) dx =
= 625650
(u) du = [ ]67
8 19
8 4 19
−2
Z
(c) (4 points)
tan4 (x) sec2 (x)dx
Solution: Let u = tan(x), du = sec2 (x) dx. Then
Z
Z
1
1
tan4 (x) sec2 (x)dx = u4 du = u5 + C = tan5 (x) + C
5
5
Z
π
3
(d) (4 points)
−π
4
sin(x)
dx
cos2 (x)
Solution: Let u = cos(x), du = − sin(x) dx. Then
Z
π
3
−π
4
Z
(e)
√
sin(x)
dx =
cos2 (x)
1
2
Z
√
2
2
−
√
1
1 1
du = [ ] 2√2 = 2 − 2
2
u
u 2
1
dx
1 − 4x2
Solution: Let u = 2x, du = 2 dx. Then
Z
Z
1
1
1
1
1
√
√
dx =
du = arcsin(u) + C = arcsin(2x) + C
2
2
2
2
2
1 − 4x
1−u
3