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Chris Phan Math 252 Spring 2005 Wednesday, 11 May 2005
Second Midterm Examination Solutions
I have rescaled the exam, so it is out of 32 points, not 35.
- Evaluate the following:
(a)
tan−^1 (x)dx (You may use (^) dxd tan−^1 (x) = (^) 1+^1 x 2 , but show all other work.) Solution: ∫ tan−^1 (x)dx = tan−^1 (x)x −
x
1 + x^2 dx
(By parts, setting u = tan−^1 (x), dv = dx, du = (^) 1+^1 x 2 dx, v = x.)
= x tan−^1 (x) −
u
du (Substitution, setting u = 1 + x^2 , du = 2xdx.)
= x tan−^1 (x) −
ln |u| + C
= x tan−^1 (x) −
ln(1 + x^2 ) + C.
The answer can also be written x tan−^1 (x) + ln
√^1 1+x^2
(b)
xexdx Solution: ∫ xexdx = xex^ −
exdx
(By parts, setting u = x, dv = exdx, du = dx, v = ex.) = xex^ − ex^ + C.
- (3 points) Suppose the height of a basketball t seconds after being thrown is given in meters by the function h(t) = − 4. 9 t^2 + 5t + 1. What is its average height in the first second after it is launched? (A decimal approximation is acceptable, but show all work used to obtain the approximation.) Solution: We want the average value of h on [0, 1]. Then
have =
0
h(t)dt
0
(− 4. 9 t^2 + 5t + 1)dt
t^3 +
t^2 + t
] 1
0
≈ 1 .8667 m.
- (6 points) Find a number c such that the region enclosed by y = x^2 − c^2 and y = c^2 − x^2 has an area of 72. Solution: We do this by finding the area of the region, in terms of c, setting equal to 72 , and solving for c.
First, we need to find x such that x^2 − c^2 = c^2 − x^2 (to get the bounds of our region):
x^2 − c^2 = c^2 − x^2 , 2 x^2 = 2 c^2 , x^2 = c^2 , x = ±c.
Note that if c is a value that makes the region have an area of 72 , then the value −c will also work, so we may assume c ≥ 0. Then the area of the region is given by ∫ (^) c
−c
(c^2 − x^2 − (x^2 − c^2 ))dx =
∫ (^) c
−c
(2c^2 − 2 x^2 )dx
= 2 c^2 x −
x^3
]c
−c = 2 c^3 −
c^3 − (− 2 c^3 +
c^3 )
=
c^3 +
c^3
=
c^3
Setting this area equal to the desired 72 , we get 8 3 c^3 = 72 ,
c^3 3
c^3 = 27 , c =
- (6 points) Consider the region bounded by y = x^2 and y =
x. What is the volume of the solid of revolution obtained by rotating this region about the x-axis? Solution: To find the bounds of this region, we note that the only two solutions of x^2 =
x are 0 , 1. (You can see this by noting that if x^2 =
x, then x^4 = x, which means 0 = x^4 −x = x(x^3 −1), which only has solutions x = 0, 1. Both of these proposed solutions satisfy the original equation.) Hence, the cross-section of the solid at x = c is a washer, with outer radius
x, and inner-radius x^2 , which means it has area A(x) = πx − πx^4. So, the volume of the solid is ∫ (^1)
0
π(x − x^4 )dx = π
0
(x − x^4 )dx
= π
x^2 2
x^5 5
] 1
0 = π(1/ 2 − 1 /5) =
3 π 10
- (5 points) Consider the region bounded by y = x
(^2) + x+5 ,^ y^ = 0,^ x^ = 1, and^ x^ = 4. Set up an integral to determine the volume of the solid of revolution obtained by rotating this region about the y-axis, using cylindrical shells, but do not evaluate. Solution: Note that for x on [1, 4], the x
(^2) + x+5 >^0 (since both numerator and denominator are positive). The area of the cylindrical shell at x has height x
(^2) + ∫ x+5^ and radius^2 πx, so the volume is given by 4 1 2 πx^
x^2 + x+5 dx.
