Math 131 Lab 3
Make sure to do at least #1 through 9(a) in Lab. Short Answers on back.
R6
R2
R6
1. Assume f is continuous and that −4 f (x) dx = 27, 0 f (x) dx = 12, and 2 f (x) dx = 20. Evaluate the following. (c)
requires a bit of thought!! Plug the endpoints into f , what numbers do you get?
Z 6
Z 0
Z 1
Z 3
a)
f (x) dx
b)
f (x) dx
c)
f (x + 1) dx
d)
x2 f (x) dx
−4
0
R6
−1
3
R6
R6
2. Suppose that −4 f (x) dx = 27, −4 g(x) dx = −12, and −4 h(x) dx = 7. Evaluate each of the following expressions
by using properties of the integal. The answer to part (b) is NOT 29.
Z 9
Z 6
Z 6
f (x − 3) dx
f (x) + 2 dx
c)
(h(x) − 4g(x)) dx
b)
a)
−1
−4
−4
3
1
3. Use this graph of f
−1
−3
Z
a)
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1
5
2
Z
f (x) dx
0
b)
3
4
5
6
7
10
8
Z
f (x) dx
c)
2
to evaluate the following.
9 10 11 12 13 14
14
Z
f (x) dx
5
5
|f (x)| dx
d)
0
e) Is the answer to (a) or (d) the Total Area between f and the x-axis on [0, 5]?
4. Evaluate these definite integrals using the Fundamental Theorem of Calculus.
Z 8
Z 4
Z 4 √
1
4 x+2
√
dx
dx
c)
a)
6x1/2 + 2x−3/2 dx
b)
3
2
x
1 2 x
1
1
Z
π/4
sec x(tan x − 2 sec x) dx
d)
0
5. Evaluate these definite and indefinite integrals. Hint: ‘Adjust.’
Z 1
Z
a)
4e−8t dt
b)
sec(3x) tan(3x) dx
0
6. These problems make use of the FTC and FTC II. Begin by determining the given derivatives.
"Z 3
#
Z x
Z 0
x
d
d
d
2
3 s
a)
cos(t + 1) dt
b)
s e ds
c)
s sin(s) ds
dx 2
dx x
dx 1
R1
d) Suppose that the graph of f (t) passes through the origin and the point (1, 2). Evaluate 0 f 0 (t) dt. Notice the
integrand is f 0 (t). So what do you get when you integrate?
Rx
e) Suppose that −1 g(t) dt = ex arcsin x. Evaluate g(0) and explain your answer. Hint: First apply FTC to determine
g.
R3
f ) Is the Fundamental Theorem of Calculus useful to calculate 1 ln(x2 ) dx? Explain.
7. In a day or two we will reverse the chain rule to create a new antiderivative process. To get into ‘shape,’ find the
derivatives of the following functions using the chain rule.
a) f (x) = e2 sin x
b) f (φ) = tan(2φ4 + 3)
c) g(t) = ln(3t + 1)
d) p(x) = (x4 + x2 )6
8. Now find these antiderivatives. Use the work you did in the previous problem and make an ‘adjustment’ when needed.
Z
Z
Z
2 sin x
2 sin x
a)
2 cos xe
dx
b)
cos xe
dx
c)
8φ3 sec2 (2φ4 + 3) dφ
Z
Z
Z
6
dt
f)
12(4x3 + 2x)(x4 + x2 )5 dx
d)
24φ3 sec2 (2φ4 + 3) dφ
e)
3t + 1
1
1
9. a) Suppose that water is slowly leaking from a still in the Chemistry Department at a rate of dv
dt = 3t+1 gals/hr.
How many gallons leak out during the time interval [1, 3]? Hint: We simply want v(3) − v(1). Translate that into
a definite integral and use work you did in the previous problem.
b) Hand in for extra credit tomorrow: If you put a 2 gallon plastic bucket under the leak (say at time 0) to catch
the water, starting at time 0 how many hours will it take for the bucket to overflow?
10. a) Suppose the function s0 (t) = 4 − t2 represents the velocity in kph of a person walking over the time interval [0, 4]
hours. How far from her starting point (displacement or net distance travelled) had she walked after 4 hours?
Hint: We simply want s(4) − s(0) where s(t) is the position at time t. (Ans: −16/3 km.)
b) How many kilometers had she walked (total distance) at the end of 4 hours? Hint: Figure out how many miles
she walked forward and how many she walked back by looking at the graph. (Ans: 16 km.)
R2
11. Properties of Integrals. Suppose that f (x) ≥ 0 on [0, 2] and f (x) ≤ 0 on [2, 5] and that 0 f (x) dx = 6 and
R5
f (x) dx = −8. Determine each of the following values. Hint: Try to make a rough sketch of such a function.
2
Z
5
Z
f (x) dx
a)
5
Z
3|f (x)| dx
c)
5
(f (x) + |f (x)|) dx
d)
0
2
0
0
5
Z
|f (x)| dx
b)
12. Functions as Models. Over the course of a year, the length of day, i.e., the number of hours of
daylight, varies.
2π
The plot is for Boston, MA with day 0 being March 21. The function y = f (x) = 3.1 sin 365
x + 12.2 is a good
description of these data where x represents time in days. Find the total amount of light over the course of a year
(365 days). (This is the same as the area under the curve below.) Use a mental adjustment.
16
14
•
•
y
•
(hr) 12
•
•
• • • • •
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
10
•
•
•
• • • •
•
•
•
8
0
20
40
60
80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
x (day)
Short Answers
Complete answers will be online.
1. 32, −5, 12, 0.
2. 55, 47, 27
3. π − 4.5, −7, 3.5, π + 4.5
4. 30, 8 + 2 ln 4, 3/2,
√
2−3
5. Multiply by the reciprocal of the constant on the “inside” of the function (this reverses the chain rule): − 21 [e−8 −
1], 13 sec(3x) + c, π3
6. a) cos(x2 + 1); (b) −x3 ex , (c) x3 sin(x3 ) · 3x2 , (d) 2; (e) HW; (f) 1, No. But see longer answer online.
7. Use the chain rule:
8. e2 sin x + c,
10. −16/3 km
d
dx [f (u)]
1 2 sin x
2e
+ c,
2 sin x
= f 0 (u) du
, 8φ3 sec2 (2φ4 + 3),
dx . 2 cos xe
tan(2φ4 + 3) + c,
3 tan(2φ4 + 3) + c,
11 −2, 14, 24, 12
3
3t+1 ,
6(4x3 + 2x)(x4 + x2 )5
2 ln |3t + 1| + c,
12. 4,453
2
2(x4 + x2 )6 + c
Math 131 Lab 3 Answers
1. a)
b)
R6
f (x) dx =
0
R0
−4
R2
f (x) dx +
R6
f (x) dx = 12 + 20 = 32.
R6
f (x) dx = −4 f (x) dx − 0 f (x) dx = 27 − 32 = −5.
0
2
R6
c) y = f (x + 1) is a horizontal shift of y = f (x) one unit to the left, so
R3
d) 3 x2 f (x) dx = 0 because the endpoints are the same.
R1
−1
f (x + 1) dx =
R2
0
f (x) dx = 12.
R6
R6
(h(x) − 4g(x)) dx = −4 h(x) dx − 4 −4 g(x) dx = 7 − 4(−12) = 55.
R6
R6
b)
= −4 f (x) dx + −4 2 dx = 27 + 2[6 − (−4)] = 47. The second integral is a rectangle with height
2 and base 10.
c) The graph of f (x − 3) on [−1, 9] is the same as the graph of f (x) on [−4, 6]—just shifted 3 units to the right. So
their areas are the same, viz., 27.
2. a)
R6
−4
R6
(f (x) + 2) dx
−4
3. The integral is ‘net area.’ So using quarter-circles and triangles,
R5
R5
R2
2
+ 21 (3)(−3) = π − 4.5
a) 0 f (x) dx = 0 f (x) dx + 2 f (x) dx = π(2)
4
R 10
R8
R 10
b) 2 f (x) dx = 2 f (x) dx + 8 f (x) dx = 12 (6)(−3) + 21 (2)(2) = −7
R 14
R8
R 10
R 12
R 14
c) 5 f (x) dx = 5 f (x) dx + 8 f (x) dx + 10 f (x) dx + 12 f (x) dx = 12 (3)(−3) + 12 (2)(2) + (2)(2) + 12 (2)(2) = 3.5
R5
R2
R5
2
d) 0 |f (x)| dx = 0 f (x) dx + 2 −f (x) dx = π(2)
+ 12 (3)(3) = π + 4.5
4
e) (d), all regions are treated as positive area.
4. Using the FTC
Z 4
4
a)
6x1/2 + 2x−3/2 dx = 4x3/2 − 4x−1/2 = (32 − 2) − (4 − 4) = 30
1
1
Z
4
4
4x−1/2 + 2x−1 dx = 8x1/2 + 2 ln |x| = (16 + 2 ln 4) − (8 + 0) = 8 + 2 ln 4
b)
1
1
Z 8
8
3
3
3
1
x−2/3 dx = x1/3 = (2 − 1) =
c) =
2 1
2
2
2
1
Z π/4
π/4
√
√
d) =
sec x tan x − 2 sec2 x dx = sec x − 2 tan x
= ( 2 − 2) − (1 − 0) = 2 − 3
0
0
5. Multiply by the reciprocal of the constant on the “inside” of the function (this reverses the chain rule):
Z 1
1
a)
4e−8t dt = − 84 e−8t = − 12 [e−8 − 1]
0
0
Z
b)
sec(3x) tan(3x) dx = 13 sec(3x) + c
R x
d
f (t) dt = f (x). So
6. FTC II says: dx
a
R x
d
a) dx
cos(t2 + 1) dt = cos(x2 + 1).
hR2
i
Rx 3 s 0 3 s
d
d
b) dx
s e ds = dx
− 0 s e ds = −x3 ex .
x
i
hR 3
R u
x
d
du
d
c) The chain rule version: dx
f
(t)
dt
=
f
(u)
.
So
s
sin(s)
ds
= x3 sin(x3 ) · 3x2
dx
dx
a
1
1
R1
d) Since f (0) = and f (1) = 2, then 0 f 0 (t) dt = f (t) = f (1) − f (0) = 2 − 0 = 2.
0
e) For Homework Later.
f ) To apply FTC I you need to know an antiderivative for the integrand.
We don’t know an antiderivative for ln(x2 ).
R
Exception: If you happen to remember Lab 1, you found that ln x dx = x ln x + c. So
Z 3
Z 3
Z 3
3
ln(x2 ) dx =
2 ln x dx = 2
ln x dx = 2[x ln x − x] = 2[(ln 3 − 3) − 1] = 2 ln 3 − 8.
1
1
1
1
3
7. Use the chain rule:
d
dx [f (u)]
= f 0 (u) du
dx .
a) f 0 (x) = 2 cos xe2 sin x
3
c) g 0 (t) =
3t + 1
b) f 0 (φ) = 8φ3 sec2 (2φ4 + 3)
d) p0 (x) = 6(4x3 + 2x)(x4 + x2 )5
8. Match these up with the answers for the previous problem and adjust with a constant where needed.
Z
Z
1
a)
2 cos xe2 sin x dx = e2 sin x + c
b)
cos xe2 sin x dx = e2 sin x + c
2
Z
Z
c)
8φ3 sec2 (2φ4 + 3) dφ = tan(2φ4 + 3) + c
d)
24φ3 sec2 (2φ4 + 3) dφ = 3 tan(2φ4 + 3) + c
Z
Z
6
dt = 2 ln |3t + 1| + c
f)
12(4x3 + 2x)(x4 + x2 )5 dx = 2(x4 + x2 )6 + c
e)
3t + 1
9. For Homework Day 8.
10. a) Net change in position s(4) − s(0) =
R4
0
s0 (t) dt =
R4
0
4
4 − t2 dt = 4t − 31 t3 0 = (16 −
64
3 )
− (0) = − 16
3 km.
b) Total distance travelled: Here we need to find the area above the x-axis and below the x-axis and treat them as
both positive and add them together. So where is 4 − t2 negative? 4 − t2 < 0 when 4 < t2 so 2 < t (or t < −2). So
we need t > 2. So we have to split the interval [0, 4] into two pieces: [0, 2] where s0 (t) is positive and [2, 4] where
s0 (t) is negative. On the first interval
Z
0
On the second interval
Z 4
2
R2
1 2
4 − t2 dt = 4t − t3 0 = (8 − 38 ) − 0 =
3
1 4
1 4
4 − t2 dt = 4t − t3 2 = 4t − t3 0 = (16 −
3
3
So the total distance covered is
R5
2
24
3
+
40
3
=
64
3
64
3 )
24
3
km.
− (8 − 83 ) = − 40
3 km.
km.
R5
f (x) dx = 6 − 8 = −2.
R5
R5
R2
b) f (x) ≤ 0 only on [2, 5] so |f (x)| = −f (x) there. So 0 |f (x)| dx = 0 f (x) dx + 2 −f (x) dx = 6 + 8 = 14.
R5
R5
c) 2 3|f (x)| dx = 3 2 −f (x) dx = 3(8) = 24.
R5
R5
R5
d) 0 (f (x) + |f (x)|) dx = 0 f (x) dx + 0 |f (x)| dx = −2 + 14 = 12.
11. a)
12.
R 365
0
0
f (x) dx =
3.1 sin
2π
365 x
0
f (x) dx +
2
+ 12.2 dx = − 365
2π · 3.1 · cos
= − 365
2π
2π
365 x
365
+ 12.2x
0
· 3.1[cos(2π) − cos 0] + (12.2 · 365 − 0) = − 365
2π [1 − 1] + 4, 453 = 4, 453 hr
4