05
Quiz 1016-273-02, 6
Name:______Solutions________
To receive full credit, you must show ALL work for each problem. This quiz is worth 30 points and covers material from Sections 7.3 and 7.4. There are two problems on the back page.
2
sec 3 ( ) = 27 u - 1 du = 27 - sec( ) + C = 9 sec 3 ( ) - 27 sec( ) + C 3
2
=9
- 27
Using x = 3 tan ( ) for original substitution
Using x = 2 sin ( ) for original substitution
=2 +
2 sin ( ) cos( ) x x 4 - x2 + C = 2 + 2 sin ( ) cos( ) + C = 2 arcsin + +C 2 2 2
2.
( 4 - x )dx = 4 (cos ( ))d = 2 (1 + cos(2 ))d = 2 + sin (22 )
2 2
9 + x2 3
3
9 + x2 1 +C = 3 3
(
)
1.
9+ x
x3
dx = 27 sec( ) tan 3 ( ) d = 27 sec 2 ( ) 1 - (sec( ) tan ( )d )
(
)
(
)
( 9+ x ) -9
2 3
9 + x2 + C
+C
3.
x2 R ( x) = S ( x) + Hint: First use long division so that where deg(R ( x) ) < deg( x - 3) . x -3 x-3
x2 x2 = + x -3 x
= x +3+
3x - (3)( x - 3) 9 9 = x +3+ = ( x + 3) + x-3 x-3 x-3
4.
2
1 3 = ln x + 2 + ln x + 6 + C 4 4
x+3 A A = 1 + 2 x + 8 x + 12 x + 2 x + 6
2
x = 2:
1 = 4 A1 + 0 - 3 = 0 - 4 A2
x = 6:
x+3 dx = x + 8 x + 12
A1 A + 2 dx = ...