Cambridge International AS and A Level Mathematics Pure Mathematics 1

Integration

P1^

6

(ii) Use your answers to part (i) to find these.

(a) (^) ∫ 42 ()xx−^3 d (b) (^) ∫()xx− 23 d
(c) (^) ∫ 72 ()xx+ 56 d (d) (^) ∫ 28 () 25 xx+^6 d
(e) (^) ∫ 62 ()xx− 1 −^4 d (f) 1
() 21 x^4
x
∫ −
d
(g) −
∫ −

4

18 x

dx (h) 8

∫ 18 − x

dx

In the activity, you saw that you can use the chain rule in reverse to integrate

functions in the form (ax + b)n.

For example,    

This tells you that (^) ∫^15 ()^32 xx+=^4 d ()^32 xc++^5
⇒ () 324 1 () 325
∫ xx+=d 15 xc++.
EXAMPLE 6.16 Find 3
∫ 52 − x
dx.
SOLUTION
3
52
35 2
(^12)
−
∫∫=− −
x
dxx() dx
Use the reverse chain rule to find the function which differentiates to give
35 2
(^12)
()− x−.
This function must be related to () 52
(^12)
− x.
Increasing the power
of the bracket by 1.
The derivative of () 52
(^12)
− x is 12 25 25 2
(^1212)
×−()−=x− −−()x−
So the derivative of − − 35 2
(^12)
()x is 35 2
(^12)
()− x−
⇒
In general, d()axdxb an()()ax b

• n n
  =+ +


• 1
  1
  Since integration is the reverse of differentiation, you can write:

⇒

d

d

() ()

()

(^325332)
1532
(^54)
4
x
x x
x

+ =× ×+

=+

35 2352

35 2

(^1212)
() ()
.

−=−− +

=− −+

−

∫ xx xc

xc

d

an ax bx ax bc

ax bx

an

n n

n

()() ()

()

()

(

++ =+ +

+=

+

∫

∫

1 +

1

1

d^1

d aax++bc).n+^1