MATH 2413 HCCS The Fundamental Theorems of Calculus in Mathematics Question

Math 2413Homework Assignment 14 (Written)
Chapter 6
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1. (10 points) Several statements about the definite integral of a continuous function f (x) over [a, b] are written below. Determine those statements that are true by writing “TRUE” and those that are false by writing “FALSE.” (No work
need be included with this question.)
Z b
I. One of the Fundamental Theorems of Calculus states that
f (x) dx = F (b)−F (a) where F 0 = f .
.
a
Z b
Z a
f (x) dx = −
II.
a
f (x) dx.
.
b
Z
III.
f (x) dx denotes all possible anti-derivatives of f (x).
Z b
IV.
.
f 0 (sin x) · cos x dx = f (sin(b)) − f (sin(a)).
.
a
V. One of the Fundamental Theorems of Calculus states that
1
d
dx
Z x
a

f (t) dt = f (x).
.
2. (10 points) A portion of the graph y = f (x) is shown in the image below, and the region bounded by this graph
and the x-axis is shaded.
Figure 1: Graph of y = f (x) over [0, 4].
Z 4
If we are told that
Z 4
f (x) dx = 16/3,
0
Z 2
f (x) dx = 37/12 and
3
Z 4
f (x) dx =
0
Z 3
f (x) dx =
2
2
f (x) dx evaluate the definite integral
2
3. (10 points) Two graphs are shown below (along with the shaded regions bounded by them and the x-axis).
A) (5 points) Calculate the deifnite integral of the graphed function without usig the Fundamental Theorem(s) of Calculus.
(Be sure to explain how you calculated this number.)
Figure 2: Graph of f (x) = 3 +
Z 7
3+
√
10x − x2 − 21.
p
10x − x2 − 21 dx =
3
B) (5 points) Calculate the deifnite integral of the graphed function without usig the Fundamental Theorem(s) of Calculus.
(Be sure to explain how you calculated this number.)
Figure 3: Graph of g(x) = |4 + 4x|.
Z 3
|4 + 4x| dx =
−2
3
4.
(10 points)
A graphZ of a function y = f (t) is shown below. Use this graph to answer the following questions
x
f (t) dt.
about the function F (x) =
−4
Figure 4: Graph of y = f (t)
A) (3 points) Compute the values of F (1) and F 0 (1).
B) (7 points) Find the maximum and minimum values of F (x) on the interval [−4, 2].
4
5. (10 points) The functions F (x) and G(x) are given by
Z x
F (x) =
e
arctan t
Z x2
dt
and G(x) =
0
earctan t dt
0
2
Note: Like the function e−t , the integrands used to define both F (x) and G(x) are not possible to anti-differentiate in terms
of our old, familiar functions, so don’t waste your precious time trying.
A) (2 points)
We

 can express the function G(x) as a composition of F (x) with another function; that is, we can write
G(x) = F h(x) . Identify the “inside function” h(x).
B) (4 points) Find F 0 (x) and G0 (x).
C) (6 points) Check that L’Hopital’s rule applies to the limits
lim
x→0
F (x)
x
and lim
x→0
and use this rule to evaluate these limits.
5
G(x)
x2
6. (40 points) Complete the following cross-word puzzle with English words (this means writing numbers out as words),
showing your work for clues 4, 7, 8, 12, 13, and 14
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Across
1. (5 points) A function f (x) with derivative f 0 (x) = ex (5 − x)3 has a local
Z 1

4. (5 points)
ln 7 · 7x dx + 1
at x = 5.
0
6. (2 points) An anti-derivative for the exponential, ex , is the
Z 8
8. (5 points)
−1
Z
9. (5 points)
√
Z e32
1 + x dx +
.
x−1 dx
1
(1 + x2 )−1 dx =
(x) + C.
11. (0 points) When computing an indefinite integral, don’t forget the
!
12. (2 points) One half of Clue 8 minus the value 15 · F (π/4) where F (x) is the function used in Clue 9.
4
14. (points) The score you will earn on your final; a.k.a. the minimum value of f (x) = 89ex + 11.
Down
2. (2 points) If a particle is traveling along a horizontal access with velocity s0 (t) = v(t) = 30t + 5 ft/sec then its
Z 0
√
√
3. (2 points) The definite integral
2 + cos x dx equals the
area under the graph 2 + cos x over [0, 2π].
2π
5. (2 points) The technique of integration that “undoes the Chain Rule” (and was also used in the first integral of Clue 8).
7. (3 points) The area under the curve of y = sec2 x over [0, π/4].
10. (5 points) If f 0 (x) > 0 over an interval, then the y-values of f (x) “
Z b
Z b
13. (2 points) The value of b that satisfies
|t| dt = 2 ·
|t| dt = 81
−9
.”
0
6
is 30 ft/sec2 .
Use this page, if needed, to include (clearly labelled) work from the previous question.
7
7. (10 points) This problem consists of two parts.
A) (7 points) Evaluate the following indefinite integral
Z
√
x2
5
ln(x + 1)
+
+
dx.
3
2
1+x
x+1
1−x
B) (3 points) Based on your work in part A), find a function y(x) that satisfies the differential equation
y 0 (x) = √
x2
ln(x + 1)
5
+
+
3
2
1+x
x+1
1−x
and initial condition y(0) = 5.
8