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Calculus II – Math 121
Test # Spring 2007
Scale: 90–100 A, 80–89 B, 70–79 C, 50–69 D. Median 82.
- 10 points each] Compute the following definite inte- grals:
a.
0
(9t^3 − 4 t^2 ) dt
Use the power rule
xn^ dx =
xn+ n + 1
∫ (^1)
0
(9t^3 − 4 t^2 ) dt =
9 t^4 4
4 t^3 3
1
0
b.
∫ (^) π/ 2
−π/ 2
(cos x − sin x) dx
Recall that the integral of cos x is sin x while the integral of sin x is − cos x.
∫ (^) π/ 2
−π/ 2
(cos x − sin x) dx
= (sin x + cos x)
π/ 2
−π/ 2 = (sin π 2 + cos π 2 ) − (− sin π 2 + cos − π 2 ) = (1 + 0) − (−1 + 0) = 2
- [10 points each] Compute the following indefinite inte- grals:
a.
∫ (^
t^2 − t−^2 t^4
dt
There is no quotient rule for integrals, so you have to simplify the integrand, then use the power rule.
∫ (^ t^2 − t−^2 t^4
dt =
(t−^2 − t−^6 ) dt
t−^1 − 1
t−^5 − 5
+ C
t
5 t^5
+ C
Be sure to always include the differential, dt in this prob- lem, until you’ve taken the integral. Once you’ve taken the integral, always include the “+C” with indefinite integrals.
b.
∫ (^ √
x √ 2
x
dx
Again, there is no quotient rule for integrals. Simplify the integrand first. ∫ (^ √ x √ 2
x
dx =
∫ (^
x^1 /^2 +
2 x−^1 /^2
dx
x^3 /^2 3 / 2
x^1 /^2 1 / 2
+ C
x^3 /^2 + 2
2 x^1 /^2 + C
c.
(x^2 + 4)(x + 1) dx
There is no product rule for integrals. Simplify the inte- grand first. ∫ (x^2 + 4)(x + 1) dx =
(x^3 + x^2 + 4x + 4) dx
= 14 x^4 + 13 x^3 + 2x^2 + 4x + C
- [10 points] Find the area of the finite region bounded between y = 4 − x^2 and y = 4 − 2 x for 0 ≤ x ≤ 2, indicated below.
The upper function is y = 4 − x^2 while the lower function is y = 4 − 2 x. Therefore, the area is ∫ (^2)
0
(4 − x^2 ) − (4 − 2 x)
dx =
0
(2x − x^2 ) dx
x^2 − 13 x^3
2
0 = (4 − 83 ]) − (0 − 0) = (^43)
- [16 points] A f (x) is drawn below. It’s graph consists of three line segments. Let P be the partition P = {− 1 , 0 , 3 , 4 }.
J J J J J J J J
JJ
y
x
a. On the graph of f (x) above, draw the rectangles corre- sponding to Lf (P ).
The partition makes three subintervals [− 1 , 0], [0, 3], and [3, 4]. On [− 1 , 0] the minimum value of the function is 0, so the first rectangle is a degenerate rectangle of height 0. On [0, 3] the minimum value of the function is 0.5, so that’s the rectangle’s height there. On [3, 4] the minimum value of the function is −1 so the rectangle has one side at y = −1.
6
J J J J J J J J
JJ
y
x
b. Compute Lf (P ), and Uf (P ).
LF (P ) is the total signed area of the rectangles you drew in part a. Except for the last rectangle, the tops are all above the x-axis, so only the last rectangle contributes a negative value.
LF (P ) = (0 − (−1))0 + (3 − 0)0.5 + (4 − 3)(−1) = 0. 5
The heights of the rectangles for UF (P ) are 1, 2, and 0.5, so
UF (P ) = (0 − (−1))1 + (3 − 0)2 + (4 − 3)0.5 = 7. 5
c. Compute
− 1
f (x) dx
It’s probably best to split the interval according to the three subintervals [− 1 , 1], [1, 2], and [2, 4]. On the first inter-
val we have
− 1
f (x) dx = 2 since it’s the area of a triangle.
On the second interval we have
1
f (x) dx = 2 since it’s
the area of a rectangle. On the third interval, either you can
integrate the function
2
(5 − 32 x) dx or you can work with
areas of triangles to get 1. Thus, the integral we’re looking for is 2 + 2 + 1 = 5.
- [12 points] Suppose f (x) and g(x) are continuous func-
tions defined on [0, 5] with
0
f (x) dx = 5,
0
f (x) dx = 2
and
3
(3g(x) + 1) dx = 20.
a. What is
5
f (x) dx?
It’s the negation of the integral from 0 to 5, namely −2.
b. What is
0
(2f (x) − 1) dx?
0
(2f (x) − 1) dx = 2
0
f (x) dx −
x
5 0
c. What is
3
g(x) dx?
We know
3
(3g(x) + 1) dx = 20, so 3
3
g(x) dx + ∫ (^5)
3
1 dx = 20. But
3
1 dx = 2, therefore 3
3
g(x) dx =
- Thus,
3
g(x) dx = 6.
d. What is
3
(f (x) + g(x)) dx?
We already know
3
g(x) dx = 6. And
3
f (x) dx =
0
f (x) dx −
0
f (x) dx = 2 − 5 = − 3 ,
so
3
(f (x) + g(x)) dx = −3 + 6 = 3.
- [12 points] The function f (x) is pictured below. Let F (x) =
∫ (^) x
0
f (t) dt be signed area under the curve from 0 to x, so
F (x) =
∫ (^) x
0
f (t) dt.
