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Calculus II
Math 116
Homework 3
Due Friday Feb. 14
(1) Let f (x) = x3 , and compute the Riemann sum of f over the interval [1, 2], using
5 subintervals of equal length and choosing the representative points to be the left
endpoints of the subintervals.
Solution. ∆x = 2−1
= 15 = .2. Then our 5 subintervals are [1, 1.2], [1.2, 1.4], [1.4, 1.6],
5
[1.6, 1.8], and [1.8, 2]. So the Riemann sum is
(.2)[f (1)+f (1.2)+f (1.4)+f (1.6)+f (1.8)] = (.2)[(1)3 +(1.2)3 +(1.4)3 +(1.6)3 +(1.8)3 ] = 3.08.
4
(2) Compute
each of the following:
R8 √
3
(a) 1 2 xdx
Solution.
Z
Z 8
√
3
2 xdx =
8
1
2x 3 dx
1
1
8
=
4
3 3
x 2
4
4
= 32 (8) 3 − 32 (1) 3 = 32 (2)4 −
3
2
=
45
.
2
1
4
(b)
Rπ
0
(ex − sin(x))dx
Solution.
π
Z π
x
x
(e − sin(x))dx = (e + cos(x)) = (eπ + cos(π)) − (e0 + cos(0))
0
0
π
= e − 1 − (1 + 1) = eπ − 3.
4
(c)
Rπ
−π
(sin(x) + cos(x))dx
Solution.
π
Z π
(sin(x) + cos(x))dx = [− cos(x) + sin(x)]
−π
−π
= [− cos(π) + sin(π)] − [− cos(−π) + sin(−π)]
= [1 + 0] − [1 + 0] = 0.
4
(d)
R2
2
dx
1 x3
Solution.
Z
1
2
2
dx =
x3
Z
1
2
2
2x−3 dx = −x−2 = [− 41 ] − [−1] = 34 .
1
4
(3) Compute
of the following using the substitution method:
R 1 2each
3
(a) −1 x (x + 1)4 dx
1
2
Solution. Let u = x3 + 1. Then du = 3x2 dx.
Z x=1
Z 1
Z
1 x=1 4
2 4 du
2 3
4
xu
x (x + 1) dx =
u du
=
3x2
3 x=−1
x=−1
−1
x=1
1
5
3
5
1
= 31 [ 15 u ]
= 15
(x + 1) −1
x=−1
1
1
((1)3 + 1)5 ] − [ 15
((−1)3 + 1)5 ] =
= [ 15
32
.
15
4
(b)
R1
ex
0 1+ex
dx
Solution. Let u = 1 + ex . Then du = ex dx.
Z 1
Z x=1 x
Z x=1
ex
e du
1
dx =
=
du
x
x
0 1+e
x=0 u e
x=0 u
x=1
x=1
= ln |u|
= ln |1 + ex |
= ln |1 + e| − ln |1 + e0 |
x=0
x=0
= ln(1 + e) − ln(2).
4
(c)
R π/2
−π/2
sin5 (x) cos(x)dx
Solution. Let u = sin(x). Then du = cos(x)dx.
Z x=π/2
Z x=π/2
du
5
5
=
u5 du
u cos(x)
sin (x) cos(x)dx =
cos(x)
x=−π/2
x=−π/2
−π/2
x=π/2
π/2
= 61 u6 = 16 sin6 (x)
Z
π/2
=
[ 61
x=−π/2
sin6 ( π2 )] − [ 16
6
sin
−π/2
π
(− 2 )] = [ 61 (1)6 ]
− [ 16 (−1)6 ] = 0.
4
(d)
R eπ
1
sin(ln(x))
dx
x
Solution. Let u = ln(x). Then du = x1 dx.
Z eπ
Z x=eπ
Z x=eπ
sin(ln(x))
sin(u)
dx =
xdu =
sin(u)du
x
x
1
x=1
x=1
x=eπ
eπ
= − cos(u)
= − cos(ln(x))
x=1
π
1
= [− cos(ln(e ))] − [− cos(ln 1))]
= [− cos(π)] − [− cos(0)] = 1 − [−1] = 2.
4
3
(4) The concentration of a certain drug in a patient’s bloodstream t hours after injection
is
0.2t
C(t) = 2
t +1
3
mg/cm . Determine the average concentration of the drug in the patient’s bloodstream over the first 4 hours after the drug injection.
Solution. We want to compute
Z
1 4 0.2t
dt,
4 0 t2 + 1
that is, we want to compute the average value of the function C(t) on the interval
[0, 4]. Let u = t2 + 1 and du = 2tdt. Then
Z
Z
Z
Z
Z t=4
1
1 t=4 0.2t du
1 t=4 .2
1 t=4 1 1
1
1 4 0.2t
dt =
=
du =
du =
du
2
4 0 t +1
4 t=0 u 2t
8 t=0 u
8 t=0 5 u
40 t=0 u
t=4
4
1
1
2
ln |u| =
ln(t + 1)
=
40
40
t=0
0
1
1
=
ln(42 + 1) −
ln(02 + 1)
40
40
1
= [ln(17) − ln(1)] = .0708mg/cm3 .
40
4
(5) Find the area bound by the given curves:
(a) y = sin(x), y = cos(x), x = π4 , and x =
5π
.
4
Solution.
5π
Z 5π
4
4
[sin(x) − cos(x)]dx = [− cos(x) − sin(x)]
π
4
π
4
= [− cos( 5π
) − sin( 5π
)] − [− cos( π4 ) − sin( π4 )]
4
4
√
√
=[
2
2
+
√
= 2 2.
2
]
2
√
− [−
2
2
√
−
2
]
2
4
(b) y = x2 and y = −x2 + 1.
Solution.
x2 = −x2 + 1 ⇐⇒ 2x2 = 1 ⇐⇒ x2 =
1
2
⇐⇒ x = ± √12 = ±
√
2
.
2
4
√
2
2
Z
√
− 22
[(−x2 + 1) − x2 ]dx =
Z
√
2
2
√
− 22
[−2x2 + 1]dx
√2
2
= [− 23 x3 + x] √
2
− 2
√
2 3
2
)
+
]
2
2
√
= [− 23 (
− [− 32 (−
√
√
√
= − 13 2 + 2 = 23 2.
√
2 3
)
2
√
−
2
]
2
4
(c) y = x2 , y =
Z
0
−1
√
3
x, x = −1, and x = 1.
Solution.
0
1
Z 1
4 4
√
√
2
2
3
3
3 3
1 3 1 3
3 3 [x − x]dx +
[ x − x ]dx = [ 3 x − 4 x ] + [ 4 x − 3 x ]
0
−1
=
=
4
[0 − ( 31 (−1)3 − 34 (−1) 3 )]
1
+ 34 + 34 − 13 = 23 .
3
+
0
4
[ 34 1 3
− 13 13 − 0]
4
2
(d) y = x and y = 4.
Solution.
x2 = 4 ⇐⇒ x = ±2.
2
Z 2
2
1 3 [4 − x ]dx = [4x − 3 x ] = [8 − 83 ] − [−8 + 83 ]
−2
= 16 −
16
3
=
−2
48−16
3
=
32
.
3
4
(6) Find the area of the region bounded above by the graph of y = x and below by the
graph of y = sin(x) from x = 0 to x = π.
Solution.
Z
π
[x − sin(x)]dx =
0
[ 21 x2
π
2
+ cos(x)] = [ π2 + cos(π)] − [cos(0)]
0
=
2
[ π2
− 1] − [1] =
π2
2
−2
4