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(MATH024)[2007](s)midterm~950^_10405.pdf
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Math 024 -Calculus II
Midterm Test, Spring Semester, 2007

Time Allowed: 2 Hours Total Marks: 55
Note: Calculators are not allowed. Please show all necessary details in your work. Only the following integrals can be used:
. n+1 .
x
x
x n dx =+ C, (n .= .1); e x dx = e + C;
n +1
dx dx
= ln |x| + C; = arctan x + C;
xx2 +1
dx
sin x dx = . cos x + C; = arcsin x + C;
. 1 . x2
cos x dx = sin x + C; . (.1)dx

= arccos x + C.
1 . x2
1. (5 Marks) Use the Fundamental Theorem of Calculus to derive the formula:
. b(x)
df(t) dt = f(b(x)) b.(x) . f(a(x)) a .(x).
dx a(x)
2. (20 Marks) Evaluate the following integrals.
2x + 1
(a) x3 . 1 dx;
.
(b) cos 4 x dx;
.
(c) ln(x + x2 + 1) dx;
.
(d) |x| dx. (to be continued to the back page)

3. (10 Marks) Explain why the following inde.nite integrals can be worked out explicitly:
.. . 1
x +1 2007
(a) dx;
x . 1
dx
(b) .
(sin x)316
4.
(5 Marks) Find the area of the 2-dimensional region enclosed by the

closed curve
|x|1/2 + |y|1/2 =1.


5.
(10 Marks)


(a) Find the area of the 2-dimensional region enclosed by the ellipse
22
xy
+ =1.
2 b2
a
(b) Use (a) to .nd the volume of the 3-dimensional region enclosed by the ellipsoid
222
xyz
++ =1.
2 b22
ac
6. (5 Marks) Show that
. 2 . lim 1+sin4 x cos(nx) dx =0.
n+
0
Solutions
1. (5 Marks) Use the Fundamental Theorem of Calculus to derive the formula:
. b(x)
df(t) dt = f(b(x)) b.(x) . f(a(x)) a .(x).
dx a(x)
Solution: Let F be an antiderivative of f, that is, F .(x)= f(x). By the Fundamental Theorem of Calculus, we have
. b(x)
f(t) dt = F (b(x)) . F (a(x)).
a(x)
Di.erentiating both sides, we have
. b(x)
df(t) dt = F .(b(x)) b.(x) . F .(a(x)) a .(x).
dx a(x)
Since F .(b(x)) = f(b(x)) and F .(a(x)) = f(a(x)), we obtain the desired
formula.
2. (20 Marks) Evaluate the following integrals.
2x +1
(a) dx;
x3 . 1
Solution: Since

1 . 1
x +
2x +1 1 x 1
= . = . . 2 .22
x3 . 1 x . 1 x2 + x +1 x . 1 13
x ++
24
thus, 2x +1
dx
x3 . 1
..2
. 1 .
dx +
11 dx
= ln |x . 1|. ..22 + ..2
2 13 2 13
x ++ x ++
24 24
.2 .
1 1 312 dt
= ln |x . 1|. ln x + ++
2 2 423 t2 +1 |x . 1| 1
= ln + arctan t + C
x2 + x +1 3
|x . 1| 1 21
= ln + arctan x ++ C
x2 + x +1 3 32
3
(b) cos 4 x dx;
Solution:
cos 4 x dx = cos 3 xd(sin x)
= cos 3 x sin x . sin x 3 cos2 x (. sin x) dx
3 22
= cos x sin x + 3 cos x (1 . cos x) dx
= cos 3 x sin x + 3 cos 2 x dx . 3 cos 4 x dx
Solving the integral from the last relation gives
13
cos 4 x dx = cos 3 x sin x + cos 2 x dx
44
1 3 3 . 1 + cos(2x)
= cos x sin x + dx
4 42
1 33
= cos 3 x sin x + x + sin(2x)+ C.
4 816
An alternative expression is
3x sin(2x) sin(4x)
+++ C.
84 32

(c) ln(x + x2 + 1) dx;
Solution:

ln(x + x2 + 1) dx
. .. = x