430 Integration (Chapter 15)Example 12 Self Tutor
Integrate with respect tox:
a 2 e^2 x¡e¡^3 x b 2 sin(3x) + cos(4x+¼)aZ
(2e^2 x¡e¡^3 x)dx=2(^12 )e^2 x¡(¡^13 )e¡^3 x+c=e^2 x+^13 e¡^3 x+cbZ
(2 sin(3x) + cos(4x+¼))dx=¡^23 cos(3x)+^14 sin(4x+¼)+cEXERCISE 15F
1 Find:aZ
(2x+5)^3 dx bZ
1
(3¡ 2 x)^2dx cZ
4
(2x¡1)^4dxdZ
(4x¡3)^7 dx eZ p
3 x¡ 4 dx fZ
10
p
1 ¡ 5 x
dxgZ
3(1¡x)^4 dx hZ
4
p
3 ¡ 4 x
dx2 Integrate with respect tox:
a sin(3x) b 2 cos(¡ 4 x)+1 c 3 cos¡x
2¢d 3 sin(2x)¡e¡x e 2 sin¡
2 x+¼ 6¢
f ¡3 cos¡¼
4 ¡x¢g cos(2x) + sin(2x) h 2 sin(3x) + 5 cos(4x) i^12 cos(8x)¡3 sinx3 Find y=f(x) given
dy
dx
=p
2 x¡ 7 and that y=11when x=8.4 The function f(x) has gradient function f^0 (x)=
4
p
1 ¡x
, and the curve y=f(x) passes through
the point (¡ 3 ,¡11).
Find the point on the graph of y=f(x) withx-coordinate¡ 8.5 Find:aZ
3(2x¡1)^2 dx bZ
(x^2 ¡x)^2 dx cZ
(1¡ 3 x)^3 dxdZ
(1¡x^2 )^2 dx eZ
4p
5 ¡xdx fZ
(x^2 +1)^3 dx6 Find:aZ ¡
2 ex+5e^2 x¢
dx bZ ¡
3 e^5 x¡^2¢
dx cZ ¡
e^7 ¡^3 x¢
dxdZ
(ex+e¡x)^2 dx eZ
(e¡x+2)^2 dx fZ μ
x¡
5
(1¡x)^2¶
dx7 Find an expression forygiven that
dy
dx
=(1¡ex)^2 , and that the graph hasy-intercept 4.=2£¡^13 cos(3x)+^14 sin(4x+¼)+ccyan magenta yellow black(^05255075950525507595)
100 100
(^05255075950525507595)
100 100 4037 Cambridge
Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_15\430CamAdd_15.cdr Monday, 7 April 2014 3:59:41 PM BRIAN