Test 1 Review Solutions - Math 113 Calculus II Test 1...

Math 113, Calculus II Test 1: Review Solutions
1. Use the limit definition of the definite integral to evaluate
5
2
(4 x - x
2
) dx.
Solution: By definition,
5
2
(4 x - x
2
) dx = lim
n →∞
∑
n
i =1
f ( x
*
i
)Δ x , where Δ x =
b - a
n
=
5 - 2
n
=
3
n
, f ( x ) = 4 x - x
2
and we may take x
*
i
= x
i
to be the right-hand endpoints of the n
subintervals, which are given as x
i
= a + i Δ x = 2 + i
3
n
= 2 +
3 i
n
. Thus,
5
2
(4 x - x
2
) dx = lim
n →∞
n
i =1
f (2 +
3 i
n
) ·
3
n
= lim
n →∞
n
i =1
[4(2 +
3 i
n
) - (2 +
3 i
n
)
2
] ·
3
n
= lim
n →∞
n
i =1
[8 +
12 i
n
- (4 +
12 i
n
+
9 i
2
n 2
)]
3
n
= lim
n →∞
n
i =1
[4 -
9 i
2
n 2
] ·
3
n
= lim
n →∞
n
i =1
[
12
n
-
27 i
2
n 3
]
= lim
n →∞
[(
12
n
n
i =1
1) - (
27
n 3
n
i =1
i
2
)]
= lim
n →∞
[
12
n
( n ) -
27
n
3
(
n ( n + 1)(2 n + 1)
6
)]
= lim
n →∞
[12 -
27
6
( n + 1)(2 n + 1)
n 2
]
= 12 -
27
6
lim
n →∞
(
n + 1
n
)(
2 n + 1
n
)
= 12 -
9
2
[(1)(2)] = 12 - 9 = 3 .
Hence,
5
2
(4 x - x
2
) dx = 3. [Here’s a quick check using FTC(II):
5
2
(4 x - x
2
) dx = 4
x
2
2
-
x
3
3
|
5
2
=
(50 -
125
3
) - (8 -
8
3
) = 42 -
117
3
= 42 - 39 = 3.]
2. There is a rule of thumb that the area of any parabolic arch is two-thirds the product of
its height h and the length of its base b . Confirm this formula by using the Fundamental
Theorem of Calculus on a suitable function to calculate the area in question.
Solution: The function f ( x ) =
4 h
b 2
( bx - x
2
) has graph which is a parabola passing through
the points (0 , 0) , ( b, 0) (so that its base is b ) and height h = f ( b/ 2). Thus we want to show
that
b
0
f ( x ) dx =
2
3
bh . So consider
b
0
4 h
b 2
( bx - x
2
) dx =
4 h
b 2
b
0
( bx - x
2
) dx =
4 h
b 2
[ b
x
2
2
-
x
3
3
]
b
0
=
4 h
b 2
[( b
b
2
2
-
b
3
3
) - (0)]
=
4 h
b 2
[ b
3
(
1
2
-
1
3
)] = 4 hb [
1
6
] =
2
3
bh
as desired.

Math 113, Calculus II Test 1: Review Solutions
3. Suppose you know that a certain function f is twice differentiable and that its graph over
[0 , 2] is given in the figure on your review sheet. As you see, the printer was sloppy and spilled
a lot of ink on the graph. Decide, if possible, whether each of the following definite integrals
is positive, equal to zero, or negative.
(a)
2
0
f ( x ) dx
(b)
2
0
f ( x ) dx
(c)
2
0
f ( x ) dx
Solution: By the Fundamental Theorem of Calculus (Part I), we know that
2
0
f ( x ) dx =
f (2) - f (0) because f ( x ) is an antiderivative of f ( x ). Now f measures the slope of the
curve, and we see that the slope of f is negative when x = 2 (because the curve is decreas-
ing there!) so that f (2) < 0. Similarly, since f is increasing when x = 0, f (0) > 0, so
f (2) - f (0), whatever the actual value of the numbers may be, must be negative as it is a
negative number ( f (2)) MINUS a positive number ( f (0)). Thus,
2
0
f ( x ) dx < 0 .
Using the Fundamental Theorem of Calculus (Part I) again,
2
0
f ( x ) = f (2) - f (0) since
f ( x ) is an antiderivative of its derivative, f ( x ). Looking at the graph of y = f ( x ), we see
that f (2) is a small positive number while f (0) is a large positive number, so that f (2) - f (0)
must then be negative. Therefore,
2
0
f ( x ) dx < 0 .
Lastly, we can see a fair bit of the graph, and the portion of it to the left of the ink spot is all
well above the x -axis and by itself contributes a positive area larger than the total area of the
ink blot BELOW the x -axis. Thus, as
2
0
f ( x ) dx measures the signed area of f ( x ) over the
interval [0 , 2] and the part of this area to the left of the ink blot is larger than any negative
amount the ink blot could be covering and the part to the right of the blot is again positive,