MTH 234 Groupwork 24 - 8.1, 8.2
(1) What is the formula for integration by parts?
Z
Z
udv = uv −
vdu
(2) When choosing what should represent u and dv in the original integral, what
are two seperate things to consider that were mentioned in lecture?
We want u to be something that becomes more simple and dv to be something that remains the same.
(3) Find the following integrals.
Z
(a)
xex dx
Here pickR u = x → du = dx and dv = ex dx → v = ex thus
= xex − ex dx = xex − x
Z
(b) (4x2 − 12)e−8x dx
(Hint: This problem requires using integration by parts twice)
Picking
1
u = 4x2 − 12, dv = e−8x ⇒ du = 8xdx, v = − e−8x
8
Z
Z
1 −8x
1 −8x
1 −8x
2
2
= (4x −12)(− e )− − e 8xdx = (4x −12)(− e )+ e−8x xdx
8
8
8
Picking
1
u = x, dv = e−8x dx ⇒ du = dx, v = − e−8x
8
Z
1
1
= (4x2 − 12)(− e−8x ) + (xe−8x − − e−8x )
8
8
1
1 1
= (4x2 − 12)(− e−8x ) + (xe−8x − · e−8x ) + c
8
8 8
1
2
Z
(c)
ln(x)dx. Check your answer by differentiating.
(Hint: Choose u = ln(x) and dv = 1)
Following the hint, choose u = ln(x) and dv = 1 ⇒ du =
1
,v = x =
x
R
R
xln(x) − x x1 dx = xln(x) − 1dx = xln(x) − x + c
Z 1
2x + 1
(d)
dx
ex
0
(Hint: Do some algebra to write this as two seperate integrals, getting
rid of the quotient. Each integral may require a different method to find
the antiderivative.)
=
R1
0
2xe−x dx +
R1
0
e−x dx, picking for the first integral
u = 2x, dv = e−x dx ⇒ du = 2dx, v = −e−x
Z
1 1 −x
1
+
e dx + [−e−x ]10 =[2xe−x ]10 − [e−x ]10 − [e−x ]10
2 0
2
3
3
1
3
=[2xe−x ]10 − [e−x ]10 = 2e−1 − (e−1 + 1) = e−1 −
2
2
2
2
√
(4) Find the average value of the function f (x) = x + x on the interval [0,4].
=[2xe−x ]10
Solution.
4
Z 4
√
1 42 2 3
1 x2 2 3
10
1
2
2
x + xdx =
=
+ x
+ (4) =
4−0 0
4 2
3
4 2
3
3
0
3
(5) An advertising firm is hired to promote a new television show. After t weeks
of the advertising campaign, it is found that P (t) percent of the the viewing
public is aware of the show, where
59t
+6
P (t) = 2
.7t + 16
(a) What is the average percentage of the viewing public that is aware of
the show during the first 5 weeks of the advertising campaign?
We are looking for the average of P (t) on the interval [0,5]. Using the
average value formula (for a function on an interval) we write:
1
=
5−0
Z
0
5
59t
+ 6 dt
.7t2 + 16
We can rewrite the integral above as the sum of 2 integrals.
1
=
5
Z
0
5
59t
2
.7t + 16
1
dt +
5
Z
5
6dt
0
The right-hand integral can be immediately evaluated using the Fundamental Theorem of Calculus.
Z
1 5
1
6dt = 6(5 − 0) = 6
5 0
5
Use the Substitution Method to evaluate the left-hand integral, and them
write out the final answer. Round to 2 decimals.
Picking = .7t2 + 16 we get du = 1.4tdt. In terms of u this translates or
integrating
and u(5) = .7(25) + 16 = 33.5
Z 5 limits to u(0) = 16
Z 33.5
1
59t
1
59 1
dt =
=
du
2
5 0 .7t + 16
5 16 1.4 u
= 8.43[ln(u)]33.5
16 = 8.43(ln(33.5) − ln(16)) = 6.23 approximately
Thus the final result is 6.23 + 6 = 12.23% approximately.
4
(b) At what time during the first 5 weeks of the campaign is the percentage
of viewers the same as the average percentage found in part (a)? Round
your answer to 2 decimal places.
59t
+ 6 ⇒ 6.23(.7t2 + 16) = 59t ⇒ 4.38t2 − 59t + 99.68 = 0
.7t2 + 16
using the quadratic formula we obtain t = 1.98 and t = 11.45 so t = 1.98
weeks approximately.
(6) Find the areas of each of the shaded regions, R1 and R2, below.
12.23 =
For R1, first find intersection points,
x2 + 1 = 7 − x ⇒ x2 + x − 6 = 0 ⇒ (x − 2)(x + 3) = 0
so x = 2, x = −3 are intersection points.
Z
2
Z
2
Z
2
(6−x−x2 )dx =
(7−x−(x +1))dx =
(7−x−(x +1)dx =
−3
−3
−3
2
3 2
2
3
2
x
x
2
2
(−3)
(−3)3
6x −
−
= 6(2) −
−
− (6(−3) −
−
) = 20.83
2
3 −3
2
3
2
3
2
R1 =
2
7 2
x2
x2 x3
For R2 =
(7−x)dx− (7−x−(x +1))dx = 7x −
− 6x −
−
2 0
2
3 0
0
0
2
3
49
2
2
49
= (49 − ) − (6(2) −
− )=
− 7.33 = 17.1667
2
2
3
2
Z
7
Z
2
2