MATH 221: Calculus I
HW 7
Summer 2013 (M. Stapf)
Due: July 23th
Instructions: (10 points) Write up all solutions on another paper. Please clearly label each question, and
circle or box your final answers.
Antidifferentiation and Indefinite Integration
Z
1
dx.
9x
1. Evaluate the indefinite integral
=
1
ln(|x|) + C
9
Z
x2 + x + 1
dx.
x
Z
1
=
x + 1 + dx
x
1
= x2 + x + ln(|x|) + C.
2
Z
√
x x dx.
2. Evaluate the indefinite integral
3. Evaluate the indefinite integral
Z
=
x3/2 dx
=
2 5/2
x +C
5
4. This question was repeated as a typo. I’m terribly sorry for all the typos in this homework.
Z
1
dx.
5. Evaluate the indefinite integral
1−x
Guess the solution looks like ln(1 − x). Thest this by differentiating.
d
1
ln(1 − x) = −
.
dx
1−x
This guess is off by a factor of −1. Correct our guess. The correct antiderivative is
= − ln(1 − x) + C.
Z
6. Determine the indefinite integral
2e3x dx by guessing the antiderivative, then choosing the correct
coefficient.
Guess a solution of e3 x.
=
2 3x
e + C.
3
7. This question was repeated as a typo.
1
Z
8. Determine the indefinite integral
√
7
dx by guessing the antiderivative, then choosing the
3x + 1
correct coefficient.
Guess a solution of
√
3x + 1.
=
14 √
3x + 1 + C.
3
9. This question was repeated as a typo.
Definite Integration
10. §6.3 Ex. 1.
2
Z
1/2
1
dx
x
11. §6.3 Ex. 2.
Z
3
x+
1
1
dx
x
12. §6.3 Ex. 3.
Z
3
(1 − x)(x − 3) dx
1
13. State the Fundamental Theorem of Calculus, and briefly explain what it means.
The Funademental Theorem of Calculus states
Z b
f (x) dx = F (b) − F (a).
a
This means we can evaluate a definite integral simply by finding the antiderivative of f (x) and evaluating
at the endpoints.
Z 1
14. Evaluate the integral
x dx.
−1
1
1 2 x 2
−1
1
1
= −
2
2
=
= 0.
Z
1
15. Evaluate the integral
x2 dx.
−1
1
1 3 x 3
−1
1
1
= − −
3
3
2
= .
3
=
2
Z
1
4ex + 3e−x dx.
16. Evaluate the integral
0
1
= 4ex − 3e−x 0
= 4e1 − 3e−1 − 4e0 − 3e0
3
= 4e − − 1.
e
Z
2
8x2 + 7 dx.
17. Evaluate the integral
−1
2
8 3
= x + 7x 3
−1
8
8
= · 8 + 14 −
(−1) − 7
3
3
64
8
=
+ 14 + + 7
3
3
= 24 + 21 = 45.
Z
ln(2)
e2x dx.
18. Evaluate the integral
0
ln(2)
1 2x = e 2
0
1 2 ln(2)
1
= e
−
2
2
3
= .
2
Z
64
19. Evaluate the integral
4
1
√ dx.
4 x
64
1 √ =
x
2
4
1√
1√
=
64 −
4
2
2
= 3.
Z
20. Evaluate the integral
1
ex + x +
√
x dx.
0
1
1
2
= ex + x2 + x3/2 2
3
0
1
2
0
1
= e + + − (e + 0 + 0)
2 3
1
=e− .
6
21. A car is travelling with a velocity given by v(t) = 30−30e−t miles per hour. Determine the displacement
after 4 hours.
3
Recall, displacement is the integral of velocity. Thus the displacement is given by
Z 4
D=
30 − 30e−t dx
0
4
= 30t + 30e−t 0
= 120 + 30e−4 − (0 + 30e0 )
= 90 + 30e−4 miles.
22. A factory produces computers at a rate of r(t) = 10 − t computers per hour over a 10 hour day.
Determine how many computers are produced during the during the time t = 2 hours and t = 6 hours.
To find the total, we will integrate the rate.
Z 6
10 − t dx
N=
2
6
1 = 10t − t2 2
2
1
1
= 60 − · 36 − (20 − · 4)
2
2
= 60 − 18 − 20 + 2
= 24.
23. Optional Challenge: Projectile is fired from the ground with an initial velocity of 30 meters per second.
The ball is subject to a gravitational acelleration of 10 meters per second squared downwards. Starting
with accelleration, the second derivative of position, x00 = a = −10, determine the equation for the
projectile’s position.
We will start with acceleration, and find the next function by integrating, and applying the given
condition.
a(t) = −10
v(t) = −10t + C1 .
Based on the given initial velocity, v(0) = 30, and so C1 = 30.
v(t) = 30 − 10t.
Now to find the position of the projectile,
h(t) = 30t − 5t2 + C2 .
Again, based on the given initial position (at the ground), h(0) = 0,and so C2 = 0. Thus the height at
time t is given by
h(t) = 30t − 5t2 .
Areas
24. §6.4 Ex. 1.
Z
2
Z
f (x) dx −
1
f (x) dx
3
4
4
25. §6.4 Ex. 2.
Z
3
f (x) − g(x) dx
2
26. §6.4 Ex. 3. Sketch figure 13, then shade region whose area is determined by the given integral.
The shaded area is the area enclosed in the first quadrant under both the red and blue functions, but
above the green function.
27. Determine both the net area and the total area under the curve for f (x) = x2 − 6x + 8 on (1, 6). For
net area, we simply have to calculate the definite integral.
Z 6
AN =
x2 − 6x + 8 dx
1
For total area, we must find where the function is positive and negative. First, we find the zeros of the
function.
0 = x2 − 6x + 8
0 = (x − 4)(x − 2).
This has zeros at x = 2 and x = 4. Checking the sign of f (x) (and noting it is an upward facing
parabola), we can see this function is below the x-axis on (2, 4). Thus we can calculate the area as
Z 2
Z 4
Z 6
AT =
x2 − 6x + 8 dx −
x2 − 6x + 8 dx +
x2 − 6x + 8 dx
1
2
4
2 4 6
1 3
1 3
1 3
x − 3x2 + 8x −
x − 3x2 + 8x +
x − 3x2 + 8x =
3
3
3
2
1
4 8
1
64
8
64
− 12 + 16 −
−3+8 −
− 48 + 24 −
− 12 + 16 + 72 − 108 + 48 −
− 48 + 24
=
3
3
3
3
3
4 4
2
= + +6+
3 3
3
28
1
=
or 9 + .
3
3
28. Determine the area of the region between the curves.
y = 2x2
y=8
x = −2
x = 2.
Sketching the functions, we will see that the region we must integrate is abobe 2x2 and below 8. Thus
we must integrate
Z 2
A=
8 − 2x2 dx
−2
2
2 = 8x − x3 3 −2
16
16
= 16 −
− −16 +
3
3
64
=
.
3
5
29. Determine the area of the region between the curves.
y = x2
y = x.
Sketching the functions, we will see the region is bounded below x and abovex2 on x = 0 to x = 1.
Z 2
x − x2 dx
A=
−2
2
1 2 1 3 = x − x 2
3 −2
1 1
= − −0
2 3
1
= .
6
30. This question was repeated as a typo.
6