MATH 152 - Recitation Quiz 1 - Spring 2015
Show all work! Do not use a calculator!
1. Evaluate the given denite integral, if it exists:
ˆ2
0
Solution: Note that f (x) =
x
(x2
2 dx
− 1)
has a vertical asymptote at x = 1. Therefore, the denite integral does
not exist. What follows is how you would evaluate the integral if it did exist:
Let u = x2 − 1. Then du = 2xdx. Therefore,
ˆ2
0
x
1
2 dx = 2
2
(x − 1)
ˆ2
0
x
(x2 −1)2
2x
1
2 dx = 2
2
(x − 1)
ˆ3
−1
3
1
1
1
1 4
2
1 1
du
=
−
·
+
1
=− · =− .
=
−
2
u
2 u −1
2 3
2 3
3
2. Evaluate the given indenite integral, if it exists:
ˆ
Solution: Let u =
ˆ
1
t.
cos (1/t)
dt
t2
Then du = − t12 dt. Therefore,
cos (1/t)
dt = −
t2
ˆ
cos (1/t)
−
dt = −
t2
ˆ
1
cos udu = − sin u + C = − sin
+ C.
t
3. Find the derivative of the given function:
ˆx
tan s2 ds
F (x) =
π
Solution: Remember that the rst part of the Fundamental Theorem of Calculus states that if
ˆx
F (x) =
f (t) dt, where c is any constant and t is some dummy variable, then F 0 (x) = f (x) . For the
c
function above, π is a constant and s is a dummy variable. So, we can make tan s2 our f (s) . Therefore,
from the Fundamental Theorem of Calculus, we know that F 0 (x) = f (x) = tan x2 .
1
4. Write (but do not evaluate) a denite integral representing the area of the region bounded
by the given curves:
y=
√
x, y =
x
.
2
Solution: First, we have to determine where the two given curves intersect. We can gure this out by setting
the two functions equal to one another and solving.
√
x=
x
x2
=⇒ x =
=⇒ 4x = x2 =⇒ 0 = x2 − 4x =⇒ 0 = x (x − 4) =⇒ x = 0 or x = 4.
2
4
Therefore, the curves intersect at exactly two points: where x = 0 and where x = 4. So we know that we will
need only one integral and that the limits of integration should be 0 and 4. All that we have to gure out
now is which function is greater on the interval [0, 4] . One way to do this is to choose a value of x between 0
and 4 and evaluate
both functions at that value of x. Since one function
involves a square root, I will choose
√
√
x = 1. Then, x = 1 and x2 = 12 . Therefore, on the interval [0, 4] , x ≥ x2 . Therefore, the following integral
gives the area of the region bounded by the two curves:
ˆ4 0
2
√
x−
x
dx.
2