Solutions to Practice Problems

Math 113, Calculus II
Test 1: Review Solutions
∫ 2a
0
Solution: We first need to figure out what region we’re talking about, namely, when is
f(x) = 0? Clearly, f(0) = 0, but if we factor
2
a 2 x − 1 a 3 x2 = x (2a − x) = 0
a3 when x = 0 or x = 2a. Thus the area enclosed by the parabola is
( 2 a 2 x− 1 a 3 x2 )dx = [ 2 a 2 (x2 2 )− 1 a 3 (x3 3 )]2a 0 = [ x2
a 2 − x3
3a 3 ]2a
0 = [( (2a)2
a 2
− (2a)3 4a2
)−(0)] =
3a3 a 2 −8a3 3a 3 = 4−8 3 = 4 3 .
13. Consider the function f(x) = x 2 on the interval [0, 3], and the region R below the graph of f
on this interval. Find:
(a) The average value f ave on [0, 3] and the number x ave in [0, 3] at which it occurs (i.e.,
f(x ave ) = f ave ).
Solution: The average value of f(x) on [0, 3] is f ave = 3−0 ∫
3 0 x2 dx = 1 3 [ x3
3 ]3 0 = 1 3 [ 27 3 ] = 3.
The value of x ave is then the number in the interval [0, 3] such that f(x ave ), which is
x 2 ave, is equal to f ave = 3. Thus all we need do is solve the equation x 2 ave = 3 and select
the solution x ave = √ 3 = 1.732 that is in [0, 3].
(b) The value of x split in [0, 3] for which the vertical line x = x split divides R into two regions
of equal area.
Solution: To find x split that divides R into two regions of equal area, we first need to
know the total area of R, which is ∫ 3
0 x2 dx = [x 3 /3] 3 0 = 9. Thus we seek to find the
value of x split in [0, 3] so that ∫ x split
0
x 2 dx = 1 2 [9] = 9 2
. Now the value of our integral is
[x 3 /3] x split
0 = 1 3 x3 split , so we must solve 1 3 x3 split = 9 2 , which is done by x3 split = 27 2
, so that
x split = ( 27 2 )1/3 = 2.3811.
14. Calculate I(x) = ∫ e −x x 5 dx.
hint: This problem will be more fun if you forego repeated integration by parts and use The
Method of Undetermined Coefficients:
Make an educated guess that
I(x) = e −x ( a 0 + a 1 x + · · · + a 5 x 5) ,
Then compute I ′ (x), set it equal to the integrand e −x x 5 and determine the coefficients
a 0 . . . a 5 .
Solution: We find I ′ (x) = [−e −x ](a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 ) + (e −x )[a 1 + 2a 2 x +
3a 3 x 2 + 4a 4 x 3 + 5a 5 x 4 ], which must be equal to d
dx [∫ e −x x 5 dx] = e −x x 5 . We observe that we
can factor e −x out of both terms in our derived version of I ′ (x), so cancelling it with the e −x
in e −x x 5 , we find we must have −(a 0 +a 1 x+a 2 x 2 +a 3 x 3 +a 4 x 4 +a 5 x 5 )+(a 1 +2a 2 x+3a 3 x 2 +
4a 4 x 3 + 5a 5 x 4 ) = (a 1 − a 0 ) + (2a 2 − a 1 )x + (3a 3 − a 2 )x 2 + (4a 4 − a 3 )x 3 + (5a 5 − a 4 )x 4 − a 5 x 5 .
Equating coefficients, we have −a 5 = 1 so a 5 = −1; 5a 5 − a 4 = 0 so a 4 = −5; 4a 4 − a 3 = 0 so
a 3 = 4(−5) = −20; 3a 3 − a 2 = 0 so a 2 = 3(−20) = −60; 2a 2 − a 1 = 0 so a 1 = 2(−60) = −120;
and a 1 − a 0 = 0 so a 0 = −120. Hence
I(x) = e −x (−120 − 120x − 60x 2 − 20x 3 − 5x 4 − x 5 ).