=========================preview======================
(MATH014)[2010](s)final~1688^_10394.pdf
Back to MATH014 Login to download
======================================================
Math014 Calculus II, Spring 2010 Final Exam Solution
1. ([9 pts]) A function f : R R has continuous derivatives up to second order, and some of its
function values are given in the following table.

x f(x) f (x)
(x)
f
Compute the following integrals: 0 -2 -1 -3

3
6 4 3

2
2 0 2
2
3f

(x)
4+[f(x)]2
0
dx [4 pts]

(a)

11
Solution Let u = f(x)such that du = f (x)dx. Then
22
2
12
f(
2
1
)
63
3f

(x)
.1
dx = du = tan

u
4+[f(x)]24(1+u2)
2

12
f(0) 0
.1
33. . 3
= tan.1 1.tan.1(.1) =+ =
2 244 4
3
(x)cosx + f(x)cosx]dx.
(b) [f

[5 pts]

0
Solution Using integration by parts,
3 3 3
[f

(x)cosx + f(x)cosx]dx
cosxdf

(x)+
f(x)cosxdx

=

0 00
/3 . /3
= f (x)cosx + f (x)sinx + f(x)cosx dx
00
/3
1 d ..
=4.(.1)+ f(x)sinx dx
2 dx
0
./3 =3+ f(x)sinx =3+6 sin =3+3 3
0 3 1
1
2. ([9 pts]) Consider the integral dx.
1+2x
0
(a) Use the trapezoidal rule with n =4 subintervals to estimate the given integral. [3 pts]
Solution By the trapezoidal rule with n =4,
1
1 111222167
dx T4 = ++++ =
1234
0 1+2x 4 2 1+2 0 1+2 1+2 1+2 1+2 120
4444
b
(b) Recall that an error bound for the estimating of f(x)dx by trapezoidal rule has the form
a
K(b.a)3
|ET| , if |f (x)| K. Determine how large should the number of intervals n
2
12n
1
1
be taken in order to guarantee that the approximation of dx by trapezoidal rule is
1+2x
0
accurate to within 0.00001. [6 pts]
(x)
Solution f(x)=(1+2x).1 , f (x)= .2(1+2x).2 and f = 8(1 +2x).3 . Note that for 0 x 1,
|f (x)= . 8 . 8
(1+2x)3
and hence
8 13

|ET|
2
12n
Choose n such that

2 200000
0.00001 .. n 258.20
2
3n3
E.g., pick n =259.

3. ([9pts]) Recallthattheprobability of a random variable(random number) X hitting a number between a and b is given by
b
P{a X b}= f(x)dx
a
if f(x)is the probability density function of X. Suppose that
. . 0 if x< 0,
.
.
f(x)= kx2(1.x) if0 x 1,
.
.
.
0 if x> 1,
where k is a constant.
(a) Find the constant k . [2 pts]
Solution
1 . 34.1
f(x)dx =1 .. kx 2(1.x)dx = kx. x= k =1
34 0 12
. 0
i.e., k =12.
(b) If F is the function de.ned by F(x)= P{X x}, .nd F(x)and sketch its graph. [5 pts]
Solution For x 0,
xx
F(x)= f(x)dx =0dx =0 .
. .
For x 1,
x 1
F(x)= f(x)dx = 12x 2(1.x)dx =1
. 0
For 0 <x< 1,
x
F(x)= 12t2(1.t)dt 1.5
0 . t4.x
t3 1
=12 .
34 0 4
=4x 3 .3x. 0.5
.
. 0 if x 0, 0
4
F(x)= 4x3 .3xif 0 <x< 1,
.
1 if x 1. -0.5
(c) Find the probability that X 1, i.e, P{X 1}. [2 pts]
33
Solution
1
1 1 418
P{X }= 12x 2(1.x)dx =1.F()=1. +=
31 327 27 9
3

4. ([10 pts]) Consider the region enclosed bythe x-axis, the vertical line x = .2, and the parametric curve de.ned by