Unit #11 - Integration by Parts, Average of a Function
Some problems and solutions selected or adapted from Hughes-Hallett Calculus.
Integration by Parts
We choose u = t
du
=1
so
dt
or du = 1 dt
1. For each of the following integrals, indicate
whether integration by substitution or integration by parts is more appropriate. Do not evaluate the integrals.
Z
(a)
x sin(x) dx
Z
x2
(b)
dx
1 + x3
Z
2
(c)
xex dx
Z
(d)
x2 cos(x3 ) dx
Z
1
√
(e)
dx
3x + 1
Z
(f)
x2 sin x dx
Z
(g)
ln x dx
and dv = sin t dt
Z
and v = sin t dt
v = − cos(t)
Using the integration by parts formula,
Z
Z
t
sin
t
dt
=
t
(−
cos
t)
−
(− cos t) dt
|{z} | {z } |{z} | {z }
| {z } |{z}
u
u
du
dv
v
v
Z
= −t cos t + cos t dt
= −t cos t + sin t + C
As always, we can check our integral is correct by differentiating:
d
(−t cos t + sin t + C) = − cos t − t(− sin t) + cos t
dt
= t sin t
= original integrand.
Z
3.
"
te5t dt
(a) This integral can be evaluated using integration by
parts with u = x, dv = sin x dx.
(b) We evaluate this integral using the substitution
w = 1 + x3 .
(c) We evaluate this integral using the substitution
w = x2 .
(d) We evaluate this integral using the substitution
w = x3 .
v = e5t /5
Using the integration by parts formula,
Z
Z
5t
5t
=
t
(e
/5)
−
(e5t /5) |{z}
dt
t
e
dt
|{z} | {z } |{z} | {z }
| {z }
(e) We evaluate this integral using the substitution
w = 3x + 1.
u
dv
u
v
v
du
Z
te5t
1
=
−
e5t dt
5
5
te5t
1 5t =
−
e /5 + C
5
5
te5t
1
=
− e5t + C
5
25
As always, we can check our integral is correct by differentiating:
d te5t
e5t
1
5e5t
−
+ C = (e5t + t(5e5t )) −
dt
5
25
5
25
5t
5t
e
e
= te5t +
−
5
5
= te5t
(f) This integral can be evaluated using integration by
parts with u = x2 , dv = sin x dx.
(g) This integral can be evaluated using integration by
parts with u = ln x, dv = dx.
To practice computing integrals by parts, do as
many of the problems from this section as you
feel you need. The problems trend from simple
to the more complex.
For Questions #2 to #25, evaluate the integral.
Z
2.
and dv = e5t dt
Z
and v = e5t dt
We choose u = t
du
so
=1
dt
or du = 1 dt
t sin t dt
= original integrand.
1
"
Using the integration by parts formula,
Z
Z
1
ln x dx = x(ln x) − x dx
x
Z
= x ln x − 1 dx
From now on, for brevity, we won’t show quite as many
steps in the solution, nor will we check the answer by
differentiating. However, you should always remember
that you can check your antiderivatives/integrals by
differentiation.
Z
4.
pe−0.1p dp
= x ln x − x + C
Z
7.
We choose u = p
and dv = e−0.1p dp
so du = dp
and v = e
−0.1p
/(−0.1)
Using the integration by parts formula,
Z
Z
p e−0.1p dp = p e−0.1p /(−0.1) − e−0.1p /(−0.1) dp
|{z} | {z } |{z} |
{z
}
|
{z
} |{z}
u
u
dv
v
v
y ln y dy
In this problem, we recall that the only simple calculus formula related to ln y is that its derivative is
d
ln(y) = 1/y. While it might be tempting
known: dy
to keep with our earlier pattern of choosing u = y and
dv = ln y dy, that won’t work because we won’t be able
to integrate dv. As a result,
du
we choose u = ln y
1
so du = dy
y
Z
pe−0.1p
+ 10 e−0.1p dp
=
−0.1
= −10pe−0.1p + 10 e−0.1p /(−0.1) + C
Z
(z + 1)e2z dz
We choose u = (z + 1)
so du = dz
and dv = e2z dz
and v = e2z /2
Using the integration by parts formula,
Z
Z
2z
2z
(z + 1)e dz = (z + 1)e /2 − e2z /2dz
Z
(z + 1)e2z
1
=
−
e2z dz
2
2
(z + 1)e2z
1 2z =
−
e /2 + C
2
2
2z
(z + 1)e
1
=
− e2z + C
2
4
Z
8.
x3 ln x dx
We choose u = ln x
1
so du = dx
x
and dv = x3 dx
and v = x4 /4
Using the integration by parts formula,
Z
Z
1
3
4
4
x ln x dx = (ln x)x /4 − (x /4)
dx
x
Z
x4 ln x 1
=
−
x3 dx
4
4
x4 ln x 1 4
=
− (x /4) + C
4
4
x4 ln x x4
=
−
+C
4
16
There is no need for further simplifications.
Z
6.
ln x dx
In this problem, we recall that the only simple calculus
formula related to ln x is that its derivative is known:
d
dx ln(x) = 1/x. This means that we have to select
u = ln x so that it will be differentiated.
We choose u = ln x
1
so du = dx
x
and v = y 2 /2
Using the integration by parts formula,
Z
Z
1
y ln y dy = (ln y)(y 2 /2) − (y 2 /2) dy
y
Z
y 2 ln y 1
=
−
y dy
2
2
y 2 ln y 1 2
− (y /2) + C
=
2
2
2
y ln y y 2
−
+C
=
2
4
= −10pe−0.1p − 100e−0.1p + C
5.
and dv = y dy
and dv = dx
Z
9.
and v = x
2
q 5 ln(5q) dq
No further simplification is necessary for a pure ‘evaluate the integral’ question.
Z
11.
x2 cos(3x) dx
5
We choose u = ln(5q)
and dv = q dq
1
(5) dq
and v = q 6 /6
so du =
5q
Using the integration by parts formula,
Z
Z
1
q 5 ln(5q) dq = (ln(5q))(q 6 /6) − (q 6 /6)
dq
q
Z
6
q ln(5q) 1
=
−
q 5 dq
6
6
q 6 ln(5q) 1 6
− (q /6) + C
=
6
6
q 6 ln(5q)
1
=
− q6 + C
6
36
Z
10.
t2 sin t dt
2
We choose u = t
so du = 2t dt
We choose u = x2
so du = 2x dx
While we have traded our original integral for a slightly
simpler one, it is still not simple enough to evaluate by
finding an obvious antiderivative. In fact, it is of the
form of one of our earlier examples of integration by
parts, so here we must apply integration by parts again
to finally evaluate the original integral.
Z
We focus on just the new integral part, x sin(3x) dx:
and dv = sin t dt
and v = − cos t
We choose u = x
so du = dx
While we have traded our original integral for a slightly
simpler one, it is still not simple enough to evaluate by
finding an obvious antiderivative. In fact, it is of the
form of one of our earlier examples of integration by
parts, so here we must apply integration by parts again
to finally evaluate the original integral.
Z
We focus on just the new integral part, t cos t dt:
so du = dt
and v = sin(3x)/3
Using the integration by parts formula,
Z
Z
x2 cos(3x) dx = x2 sin(3x)/3 − (sin(3x)/3) (2x dx)
Z
x2 sin(3x) 2
=
−
x sin(3x) dx
3
3
Using the integration by parts formula,
Z
Z
2
2
t sin t dt = −t cos t − (− cos t) (2t dt)
Z
= −t2 cos t + 2 t cos t dt
We choose u = t
and dv = cos(3x) dx
and dv = sin(3x) dx
and v = − cos(3x)/3
Using the integration by parts formula,
Z
Z
x sin(3x) dx = −x cos(3x)/3 − (− cos(3x)/3) dx
Z
x cos(3x) 1
+
cos(3x) dx
=−
3
3
x cos(3x) 1
=−
+ (sin(3x)/3) + C
3
3
x cos(3x) 1
+ sin(3x) + C
=−
3
9
and dv = cos t dt
and v = sin t
Using the integration by parts formula,
Z
Z
t cos t dt = t sin t − (sin t) dt
Z
= t sin t − sin t dt
Subbing this result back into the original integral,
Z
x2 cos(3x) dx
Z
x2 sin(3x) 2
=
−
x sin(3x) dx
3
3
|
{z
}
Second by parts step
|
{z
}
First by parts step
x2 sin(3x) 2
x cos(3x) sin(3x)
=
−
−
+
+C
3
3
3
9
x2 sin(3x) 2
2 sin(3x)
=
+ x cos(3x) −
+ C2
3
9
27
= t sin t − (− cos t) + C
= t sin t + cos t + C
Subbing this result back into the original integral,
Z
Z
t2 sin t dt = −t2 cos t + 2
t cos t dt
| {z }
Second by parts step
|
{z
}
First by parts step
where C2 = −(2/3)C is a new integration constant.
= −t2 cos t + 2 (t sin t + cos t) + C
3
Z
12.
Applying integration by parts again to the integral
marked I2 will lead to
(ln t)2 dt
I2 =
We only know how to differentiate (ln t)2 , so we have
to choose it as u.
We choose u = (ln t)2
2 ln t
dt
so du =
t
so the overall integral will be
Z
Z
t2 e5t
2
t2 e5t dt =
−
te5t dt
5
5
| {z }
and dv = dt
and v = t
I2
Using the integration by parts formula,
Z
Z
2 ln t
dt
(ln t)2 dt = (ln t)2 t − (t)
t
Z
= t(ln t)2 − 2 ln t dt
t2 e5t
2 te5t
e5t
=
−
−
+C
5
5
5
25
t2 e5t
2te5t
e5t
=
−
+2
+ C2
5
25
125
where C2 is a multiple of the original C.
Z p
14.
y y + 3 dy
We are left with a simpler integral, but not an easy
one (unless you look at the examples from the course
notes!)
Z
We focus on I2 =
ln t dt, and apply integration by
We choose u = y and dv = (y + 3)1/2 dx,
2
so du = dx and v = (y + 3)3/2 :
3
Z
Z p
2
2
3/2
(y + 3)3/2 dy
y y + 3 dy = y(y + 3) −
3
3
2
2 (y + 3)5/2
= y(y + 3)3/2 −
+C
3
3
5/2
4
2
= y(y + 3)3/2 − (y + 3)5/2 + C.
3
15
Z
√
15. (t + 2) 2 + 3t dt
parts once more.
We choose u = ln t
1
so du = dt
t
and dv = dt
and v = t
Using the integration by parts formula,
Z
Z
1
ln t dt = (ln t)t − (t)
dt
t
Z
= t(ln t) − 1 dt
= t ln t − t + C
√
Let u = t + 2 and dv = 2 + 3t,
so du = dt and v = 92 (2 + 3t)3/2 .
Thus, going back to our original integral,
Z
Z
(ln t)2 dt = t(ln t)2 − 2 ln t dt
| {z }
Then
Z
Z
√
2
2
(t + 2) 1 + 3t dt = (t + 2)(2 + 3t)3/2 −
(2 + 3t)3/2 dt
9
9
2
= (t + 2)(2 + 3t)3/2
9
4
−
(2 + 3t)5/2 + C.
135
Z
16. (p + 1) sin(p + 1) dp
I2
2
= t(ln t) − 2(t ln t − t + C)
Z
13.
t2 e5t dt
We choose u = t2
so du = 2t dt
e5t
te5t
−
+C
5
25
and dv = e5t dt
and v = e5t /5
Let u = p + 1 and dv = sin(p + 1),
so du = dx and v = − cos(p + 1).
Z
(p + 1) sin(p + 1) dp
Z
= − (p + 1) cos(p + 1) + cos(p + 1) dp
Using the integration by parts formula,
Z
Z
2 5t
2 5t
t e dt = t e /5 − (e5t /5)(2t dt)
Z
t2 e5t
2
−
te5t dt
=
5
5
| {z }
= − (p + 1) cos(p + 1) + sin(p + 1) + C
I2
4
Z
17.
I1 can be evaluated using integration by parts.
1
Let u = t and dv = √5−t
dt,
z
dz
ez
so du = dx and v = −2(5 − t)1/2 .
Z
t
√
dt
5−t
Rewriting the integral,
Z
Z
z
dz = ze−z dz
ez
1/2
= −2t(5 − t)
Let u = z, dv = e−z dx.
Thus du = dz and v = −e−z . Integration by parts
gives:
Z
Z
ze−z dz = −ze−z − (−e−z ) dz
Z
−z
= −ze + e−z dz
Z
Adding the two integrals back together, we obtain
Z
4
t+7
√
dt = −2t(5 − t)1/2 − (5 − t)3/2 + C
3
5−t
|
{z
}
Let u = ln x, dv = x12 dx.
Then du = x1 dx and v = − x1 .
Z
19.
√
I1
+ −14(5 − t)1/2 + C1
|
{z
}
I2
1 1
−
dx
x x
1
dx
x2
Z
21.
+C
√1 ,
5−y
so du = dy and v = −2(5 − y)1/2
Z
Z p
1/2
y 5 − y dy = −2y(5 − y) − (−2)(5 − y)1/2 dy
Z
= −2y(5 − y)1/2 + 2 (5 − y)1/2 dy
I2
This second integral, I2 , can be evaluated with a second application of integration by parts. This was done
earlier in Question #7 (using y instead of x though):
2
= −2y(5 − y)1/2 + 2 (5 − y)3/2 (−1) + C
3
4
1/2
= −2y(5 − y) − (5 − y)3/2 + C
3
Z
20.
Z
x ln x dx =
{z
}
|
x2 ln x x2
−
+C
2
4
I2
Going back to the original integral
Z
x(ln x)2 dx
2
1
x ln x x2
= x2 (ln x)2 −
−
+C
2
2
4
|
{z
}
t+7
√
dt
5−t
Since we have a fraction in the numerator, we can split
the integral into a sum, and then evaluate each term
separately.
Z
Z
Z
t+7
t
7
√
√
dt =
dt + √
dt
5−t
5−t
5−t
|
{z
} |
{z
}
I1
x(ln x)2 dx
Select u = (ln x)2 and dv = x dx,
1
so du = 2 ln x
dx and v = x2 /2.
x
Using the integration by parts formula,
2 Z 2
Z
x
x
2 ln x
x(ln x)2 dx = (ln x)2
−
dx
2
2
x
Z
1
= x2 (ln x)2 − x ln x dx
2
|
{z
}
y
dy
5−y
Let u = y and dv =
(5 − t)1/2 dt
= −14(5 − t)1/2 + C1
ln x
dx
x2
Integrating by parts, we get:
Z
Z
ln x
1
dx
=
−
ln
x
−
x2
x
Z
1
= − ln x +
x
1
1
= − ln x −
x
x
+2
4
= −2t(5 − t)1/2 − (5 − t)3/2 + C.
3
I2 can be integrated directly:
Z
7
√
dt = 7(2)(5 − t)1/2 (−1) + C1
5−t
= −ze−z − e−z + C
18.
Z
I2
1
1
x2
= x2 (ln x)2 − x2 ln x +
+C
2
2
4
I2
5
1
dw = dx:
98x
Z
Z
7x
1
7x
dx
=
dw
1 + 49x2
w 98x
|
{z
}
Z
22.
Let w = 1 + 49x2 , so
arcsin(w) dw
We don’t know the integral of arcsin, but we do know
its derivative. Therefore we pick
u = arcsin(w) and dv = dw,
1
dw and v = w.
so du = √
1 − w2
Using the integration by parts formula,
Z
Z
√
arcsin(w) dw = w arcsin(w) −
|
dw
dx
= 98x or
I2
1 1
dw
14 w
1
ln |w| + C
=
14
1
=
ln |1 + 49x2 | + C
14
=
w
dw
1 − w2
{z
}
Going back to the original integral,
Z
Z
7x
dx
arctan(x) dx = x arctan(x) −
1 − 49x2
|
{z
}
I2
The new integral I2 can be evaluate using a substitution.
−1
dz
Let z = 1 − w2 , so dw
dz = dw:
= −2w or
2w
Z
Z
w
w −1
√
√
dw =
dz
z 2w
1 − w2
|
{z
}
I2
1
= x arctan(x) −
ln |1 + 49x2 | + C2
14
Z
x arctan(x2 ) dx
24.
I2
−1 1
√ dz
2 z
−1
=
(2z 1/2 ) + C
2p
= − 1 − w2 + C
This question starts off best with a substitution, due
to the x2 inside the arctan, and the x outside:
1
Let w = x2 , so 2x
dw = dx
=
Z
1
dw
x arctan(w)
2x
Z
1
=
arctan(w) dw
2
|
{z
}
x arctan(x2 ) dx =
Going back to the original integral,
Z
Z
arcsin(w) dw = w arcsin(w) −
√
|
w
dw
1 − w2
{z
}
I2
Evaluating I2 , we use by parts, following the same approach as Question #23, but without the ‘7’ factor,
Z
1
arctan(w) dw = w arctan(w) − ln |1 + w2 | + C
2
|
{z
}
I2
p
= w arcsin(w) − − 1 − w2 + C
p
= w arcsin(w) + 1 − w2 + C2
I2
Z
23.
Z
arctan(7x) dx
Going back to the original integral, and using w = x2 ,
Z
x arctan(x2 ) dx
Z
1
=
arctan(w) dw
2
|
{z
}
We don’t know the integral of arctan, but we do know
its derivative. Therefore we pick
u = arctan(7x) and dv = dx,
7
so du =
dx and v = x.
1 + (7x)2
Using the integration by parts formula,
Z
I2
1
1
w arctan(w) − ln |1 + w2 | + C
2
2
1
1
=
x2 arctan(x2 ) − ln |1 + (x2 )2 | + C
2
2
1 2
1
= x arctan(x2 ) − ln |1 + x4 | + C2
2
4
=
Z
arctan(7x) dx = x arctan(7x) −
7x
dx
1 − 49x2
|
{z
}
I2
The new integral I2 can be evaluate using a substitution.
6
Z
25.
3
Z
2
x3 ex dx
t ln(t) dt
29.
1
2
In this problem, we note that we can’t integrate ex by
itself (no closed-form anti-derivative). However, if we
package it with one of the x’s from the x3 , we’ll get
2
xex , and that can be integrated, using substitution
(w = x2 ).
The integral is the same as in Question #7. We use by
parts, with u = ln(t) and dv = t dt.
3
Z
1
x2
2
t2 ln t t2 3
− | 2 {z 4} 1
from #7
9 ln 3 9
1 ln 1 1
=
−
−
−
2
4
2
4
9 ln 3
=
− 2 ≈ 2.944
2
t ln(t) dt =
Let u = x and dv = xe ,
2
so du = 2x dx and v = 12 ex . Using that,
Z
Z
2
2
1
1 2
1 2 x2
3 x2
x e dx = x e − xex dx = x2 ex − ex + C
2
2
2
| {z }
I2
where I2 is evaluated using the same substitution w =
x2 .
Z 5
26.
ln t dt
We follow the work from Question #23, but without
the ‘7’ factor, using u = arctan(y) and dv = dy.
We integrate by parts, as done in Question #6:
Z 5
5
ln t dt = t ln t − t {z
}
|
1
1
from #6
Z
0
= (5 ln 5 − 5) − (1 ln 1 − 1)
= 5 ln 5 − 4.
27.
arctan(y) dy
0
1
Z
1
Z
30.
1
1
1
arctan(y) dy = y arctan(y) − ln |1 + y 2 |
2
0
1
= 1 · arctan(1) − ln(2) − 0 · arctan(0) −
2
π 1
= − ln(2) ≈ 0.439.
4
2
5
x cos x dx
3
5
Z
31.
We use the solution to Question #6, or applying by
parts with u = ln(1 + t) and dv = dt.
3
3
ln(1 + t) dt
0
Integrating by parts with u = x and dv = cos x dx
gives
Z 5
5
x cos x dx = x sin(x) + cos(x)
5
Z
= 5 sin(5) + cos(5) − (3 sin(3) + cos(3))
0
≈ −3.944
5
ln(1 + t) dt = (1 + t) ln(1 + t) − (1 + t) {z
} 0
|
from #6, with x=1+t
= (6 · ln 6 − 6) − (1 · ln 1 − 1)
Z
28.
= 6 ln 6 − 5 ≈ 5.751
10
ze−z dz
0
Z
32.
The integral is the same as in Question #17. We use
by parts with u = z and dv = e−z dz, giving
Z 10
10
ze−z dz = −ze−z − e−z |
{z
}
0
0
from #17
1
arcsin z dz
0
Using the solution from Question #22, or u = arcsin z
and dv = dz,
Z 1
1
p
arcsin z dz = z arcsin(z) + 1 − z 2 0
0
√ √ = arcsin(1) + 0 − 0 · arcsin(0) + 1
π
= − 1 ≈ 0.571
2
= −10e−10 − e−10 − (0 − e0 )
= −11e−10 + 1 ≈ 0.9995
7
Z
To evaluate the integral, we follow the work from
Question #23, but without the ‘7’ factor, using u =
arctan(z) and dv = dz.
Z 2
2
1
arctan(z) dz = z arctan(z) − ln |1 + z 2 |
2
0
0
1
1
= 2 arctan(2) − ln 5 − 0 arctan(0) −
2
2
ln 5
= 2 arctan(2) −
2
1
33.
x arcsin(x2 ) dx
0
We first simplify the integral with the substitution
w = x2 , which leads to the new limits
x = 0 → w = 02 = 0 and
x = 1 → w = 12 = 1.
Z x=1
Z
1 w=1
x arcsin(x2 ) dx =
arcsin(w) dw
2
x=0
| w=0 {z
}
after substitution
36. Z
Use integration
ex sin(x) dx.
At this point, we have returned to the integral in Question #22, which can be evaluated using by parts, with
u = arcsin(w) and dv = dw.
Z x=1
x arcsin(x2 ) dx
Z
1 w=1
arcsin(w) dw
=
2 w=0
1
p
1
=
w arcsin(w) + 1 − w2 2|
{z
} 0
by parts
√ √ i
1 h
=
arcsin(1) + 0 − 0 · arcsin(0) + 1
2
π 1
= − ≈ 0.285
4
2
twice
to
find
To evaluate I2 , we select the trig function again as
u and the exponential as dv: Let u = cos(x) and
dv = ex dx,
so du = − sin(x) dx and v = ex .
Z
ex sin(x) dx
|
{z
}
our goal, I
Z
= sin(x)ex − cos(x)ex dx
|
{z
}
34. Find the area under the curve y = te−t on the
interval 0 ≤ t ≤ 2.
The function te−t is always positive on the interval
0 ≤ t ≤ 2 so the area under the curve is equal to
the integral
Z
I2
2
te
parts
There are several ways to evaluate this integral; we’ll
show just one here.
Let u = sin(x) and dv = ex dx,
so du = cos(x) dx and v = ex .
Z
Z
ex sin(x) dx = sin(x)ex − cos(x)ex dx
{z
}
|
{z
}
|
I2
our goal, I
x=0
−t
by
x
x
Z
x
= sin(x)e − cos(x)e − (− sin(x))e dx
Z
x
x
x
= sin(x)e − cos(x)e − (− sin(x))e dx
dt
0
Proceeding in the same way as Question #17, using
u = t and dv = e−t dt,
Z 2
2
te−t dt = −te−t − e−t {z
} 0
|
0
from #17
= −2e−2 − e−2 − 0e0 − e0
Tidying, we obtain
Z
Z
x
x
x
e sin(x) dx = sin(x)e − cos(x)e − sin(x)ex dx
|
{z
}
|
{z
}
= −3e−2 + 1
I
I
Grouping the integrals I, which are what we are looking for,
Z
2 ex sin(x) dx = sin(x)ex − cos(x)ex
Z
1
or
ex sin(x) dx = (sin(x)ex − cos(x)ex )
2
35. Find the area under the curve f (z) = arctan z
on the interval [0, 2].
On the interval t ∈ [0, 2], the function arctan(z)
is
Z always positive, so the area equals the integral
2
arctan(z) dz.
0
8
37. Z
Use integration
ey cos(y) dy.
by
parts
twice
to
We are simply asked to change one integral into another, which can be done here directly with integration
by parts.
Let u = xn and dv = cos(ax) dx,
1
so du = nxn−1 dx and v = sin(ax)
a
Applying the by parts formula,
find
This question is done in the same manner as the previous one. For variety, and to show it works as well, we
will select the exponential function as u and the trig
functions as dv. (Both choices work, so long as you are
consistent in both integration by parts steps.)
xn cos(ax) dx
Z
1
1
n
sin(ax) −
sin(ax) · n · xn−1 dx
=x
a
a
Z
n
1
n
xn−1 sin(ax) dx
=x
sin(ax) −
a
a
Let u = ey and dv = cos(y) dy,
so du = ey dy and v = sin(y).
Z
Z
y
y
e cos(y) dy = sin(y)e − sin(y)ey dy
{z
}
|
{z
}
|
I2
our goal, I
which is the desired formula.
39. The concentration, C, in ng/ml, of a drug in
the blood as a function of the time, t, in hours
since the drug was administered is given by
To evaluate I2 , we select the exponential function again
as u and the trig as dv:
Let u = ey and dv = sin(y) dy,
so du = ey dy and v = − cos(y).
Z
ey cos(y) dy
|
{z
}
our goal, I
Z
y
= sin(y)e − sin(y)ey dy
|
{z
}
C = 15te−0.2t .
The area under the concentration curve is a
measure of the overall effect of the drug on
the body, called the bioavailability. Find the
bioavailability of the drug between t = 0 and
t = 3.
We have
I2
Z
= sin(y)ey − (− cos(y))ey − (− cos(y))ey dy
Z
= sin(y)ey + cos(y)ey − cos(y)ey dy
Z
15te−0.2t dt.
0
We first use integration by parts to evaluate the indefinite integral of this function.
Let u = 15t and dv = e−0.2t dt,
so du = 15 dt and v = −5e−0.2t . Then,
Tidying, we obtain
Z
Z
ey cos(y) dy = sin(y)ey + cos(y)ey − cos(y)ey dy
|
{z
}
|
{z
}
I
3
Bioavailability =
Z
−0.2t
15te
I
Grouping the integrals I, which are what we are looking for,
Z
2 ey cos(y) dy = sin(y)ey + cos(y)ey
Z
1
or
ey cos(y) dy = (sin(y)ey + cos(y)ey )
2
Z
) − (−5e−0.2t )(15 dt)
Z
= −75te−0.2t + 75 e−0.2t dt
dt = (15t)(−5e
−0.2t
= −75te−0.2t − 375e−0.2t + C.
Z
Thus,
0
3
3
15te−0.2t dt = (−75te−0.2t − 375e−0.2t )
0
= −329.29 + 375 = 45.71.
38. Use integration by parts to show that
Z
Z
1
n
xn cos(ax) dx = xn sin(ax)−
xn−1 sin(ax) dx
a
a
The bioavailability of the drug over this time interval
is 45.71 (ng/ml)-hours
9
Then du = dt, and v = (−1/a)e−at .
40. During a surge in the demand for electricity, the
rate, r, at which energy is used (in megawatts)
can be approximated by
Z
E=
T
te−at dt
0
r = te−at
= −(t/a)e
−
0
where t is the time in hours and a is a positive
constant.
−aT
= −(1/a)T e
−(1/a)e−at dt
0
Z
T
e−at dt
+ (1/a)
= −(1/a)T e−aT + (1/a2 )(1 − e−aT ).
(b)
(b) What happens to E as T → ∞?
lim E = −(1/a) lim
T →∞
T →∞
T
eaT
2
+ (1/a ) 1 − lim
T →∞
1
eaT
Since a > 0, the second limit on the right hand side in
the above expression is 0. In the first limit, although
both the numerator and the denominator go to infinity, the denominator eaT goes to infinity more quickly
than T does (can verify with l’Hopital’s rule). So in
the end the denominator eaT is much greater than the
T
numerator T . Hence lim aT = 0.
T →∞ e
Thus limT →∞ E = a12 .
We know that dE
dt = r, so the total energy E used in
the first T hours is given by
T
te−at dt.
E=
T
0
(a) Find the total energy, E, in megawatt hours
used in the first T hours. Give your answer
as a function of a.
Z
Z
T
−at 0
In words this means that the total amount of energy in
1
the surge, accounting for over all time (T → ∞) is 2
a
megawatt hours.
We use integration by parts.
Let u = t, dv = e−at dt.
Average Value
41. (a) Using the graph
Z 6
f (x) dx.
shown
below,
(b) The average value is a vertical quantiy, or (total
area) divided by (total horizontal length):
find
1
f (x)
3
2
1
x
0
0
1
2
3
4
5
6
(b) What is the average value of f on [1, 6]?
1
6−1
The integral represents the area below the graph of
f (x) but above the x-axis.
(a) Since each square has area 1, by counting squares
and half-squares we find
Z
6
f (x) dx = 8.5.
1
10
Z
1
6
8.5
f (x) dx =
= 1.7
5
42. (a) Using the
Z 3
f (x) dx.
graph
below,
(ii) Similarly,
estimate
−3
Average(g) =
(b) Which of the following average values of
f (x) is larger?
(i) Between x = −3 and x = 3, or
(ii) Between x = 0 and x = 3?
1
2−0
2
Z
g(x) dx
=
0
1 1
1
· =
2 2
4
(iii) Since f (x) is nonzero only for 0 ≤ x < 1 and g(x)
is nonzero only for 1 < x ≤ 2, the product f (x)g(x) =
0 for all x. Thus
f (x)
4
3
2
1
Average(f · g) =
2−0
1
Z
2
1
=
2
f (x)g(x) dx
0
Z
2
0 dx = 0.
0
x
−4 −3 −2 −1
−1
1
2
3
4
(b) Since the average values of f (x) and g(x) are
nonzero, their product is nonzero. Thus the left side
of the statement, Avg(f )·Avg(g), is nonzero. However,
the average of the product f (x)g(x) is zero. Thus, the
right side of the statement is zero, so the statement is
not true.
−2
−3
−4
(a) The integral is the area above the x-axis minus
the area below the x-axis. Thus, we can see that
Z
44. For the function f below, write an expression
involving one or more definite integrals of the
original f (x) which would denote:
3
f (x) dx is about −6 + 2 = −4
−3
(the negative of the area from t = −3 to t = 1 plus
the area from t = 1 to t = 3.)
(b) Since the integral in part (a) is negative, the average value of f (x) between x = −3 and x = 3 is
negative. From the graph, however, it appears that
the integral of f (x) from x = 0 to x = 3 is positive
overall, meaning that the average value will also be
positive. Hence (ii) is the larger quantity.
(a) The average value of f for 0 ≤ x ≤ 5.
(b) The average value of |f | for 0 ≤ x ≤ 5.
f (x)
x
1
2
3
4
5
43. (a) Using the graphs of f (x) and g(x) shown
below, find the average value on 0 ≤ x ≤ 2
of
(i) f (x)
(ii) g(x)
(iii) f (x) · g(x)
f (x)
g(x)
1
1
1
(a) (Average value of f on [0, 5]) =
5
x0
0
0
1
2
Z
5
f (x) dx.
0
x
0
1
(b) (Average value of |f | on [0, 5]) requires that we
only keep the value of f (x) as a positive contribution to the overall integral. Since the function
switches to negative f (x) values on x ∈ [2, 5], we
have to subtract the contributions of the integral
on that interval (making them positive overall).
2
(b) Is the statement that
“Average(f ) · Average(g) =
Average(f · g)”
true or not? Explain your answer.
1
=
5
(i) Since the triangular region under the graph of f (x)
1
has area , we have
2
Z 2
1
1 1
1
Average(f ) =
f (x) dx
= · =
2−0 0
2 2
4
=
11
1
5
Z
5
|f (x)| dx
0
Z
2
Z
f (x) dx −
0
2
5
f (x) dx .
(a) Since f (x) = sin x over [0, π] is between 0 and 1,
the average of f (x) must itself be between 0 and
1. Furthermore, since the graph of f (x) is concave down on this interval, the average value must
be greater than the average height of the triangle
shown in the figure below, namely y = 0.5.
45. For the same function f in the previous question,
consider the average
value of f over the following intervals:
I. 0 ≤ x ≤ 1
II. 0 ≤ x ≤ 2
III. 0 ≤ x ≤ 5 IV. −2 ≤ x ≤ 2
(a) For which interval is the average value of f
least?
(b) For which interval is the average value of f
greatest?
(c) For which pair of intervals are the average
values equal?
(b) Average =
We’ll show that in terms of the average value of f ,
1
π−0
Z
π
sin x dx ≈ 0.64.
0
47. (a) √
What is the average value of f (x) =
1 − x2 over the interval 0 ≤ x ≤ 1? (Hint:
we don’t have the integration tools to evaluate the integral of f (x) from first principles, but f (x) describes the graph of a common geometric shape whose area we already
know.)
I > II = IV > III
Using the definition of average value and the fact that
f is even, we have
Z 2
1
f (x) dx
2−0 0
Z
1 1 2
f (x) being even: =
f (x) dx
2 2 −2
Z
1 2
multiplying, =
f (x) dx
4 −2
Average value of f on II =
(b) How can you tell whether this average value
is more or less than 0.5 without doing any
calculations?
2
2
(a) Recall
√ that a circle is defined by x + y = 1, so
2
y = 1 − x is the graph of the top half of the unit
circle. Therefore, the average value of
Wait, that’s just... = Average value of f on [−2, 2]
= Average value of f on interval IV.
f (x) =
so the average on II ([0, 2]) = average on IV ([-2, 2]).
Since f is decreasing on [0,5], the average value of f on
some interval [0, c], where 0 ≤ c ≤ 5, is decreasing as
a function of c. The larger the interval the more low
values of f are included, lowering the averag e. Hence
p
1 − x2 =
1
1−0
Z
1
p
1 − x2 dx
0
= (area of a quarter circle with r = 1)
π
= ≈ 0.79.
4
(b) The area between the graph of y = 1 − x √
and the
x-axis is 0.5. Because the graph of y = 1 − x2
is concave down, it lies above the straight line
y = 1 − x, so its average value is above 0.5. See the
figure below.
Average value of f on [0, 1]
> Average value of f on [0, 2]
> Average value of f on [0, 5]
or avg on I > avg on II, IV > avg on III
46. (a) Without computing any integrals, explain
why the average value of f (x) = sin x on
[0, π] must be between 0.5 and 1.
(b) Compute the exact average of sin x on [0, π].
12