Catholicjunior College

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CATHOLICJUNIOR COLLEGE

ADDITIONAL MATHEMATICS 4018

AO MATHEMATICS 8174

INTEGRATION  STANDARD FORMULA

SUMMARY

The process of finding an expression for y in terms of x from the gradient function, , is called integration.

It reverses the operation of differentiation.

 Adx = Ax + C

 Axn dx= xn + 1 + C

 (Ax + B)n dx = + C

ASSIGNMENT

1. Evaluate the following:

a) dx[x4 + C]b) dx[ 5x + C]

c) dx[x3/2 + C]d) dx[ + C]

e) dx[4 + C]f) dx[ + C]

2. Integrate the following with respect to x:

a) 6x + 3[3x2 + 3x + C]b) 4[4x + C]

c) 3x(x + 2)[x3 + 3x2 + C]d) (x 1)(x + 2)[x3 + x2 2x + C]

e) x(2 + )[x2 + x + C]f) [2x+ C]

g) x2 + [x3+ C]h) [x+ C]

i) 3 [3xx3/2 + C]j) (+ 3)[x2 + 2x3/2 + C]

k) ax + b[ax2 + bx + C]l) abx2[axx3 + C]

m) 2 + 4x 3x2[2x + 2x2x3 + C]n) (x4)[x5 + + C]

o) (2x)2[x3x5/2 + x2 + C]p) [x3/2 + 2x1/2 + C]

CATHOLICJUNIOR COLLEGE

ADDITIONAL MATHEMATICS 4018

AO MATHEMATICS 8174

INTEGRATION Definite Integrals

SUMMARY

Suppose f(x) is the integrand and F(x) is the anti-derivative of f(x).

Then, the definite integral of f(x) between two limits x = a and x = b is given by:

=

= F(b)  F(a).

Some properties of Definite Integrals

a) = 0

b) = 

c) = k where k is a constant

d) = +

e) + =

ASSIGNMENT

1. Evaluate the following definite integrals:

a) dx[]b) dx[]

c) dx[]d) dx[0]

e) dx[]f) dx[]

g) dt[]h) dt[]

i) dr[]j) dr[]

2. Find the value of k if dx = 1[k =  5]

3. Show that () = . Hence, or otherwise, evaluate dx.[]

4. Show that = . Hence, evaluate dx.[, ]

5. Given that y = x, show that = . Hence, evaluate dx.[6]

6. Given that dx = 8 and dx = 3, find a) dx b) dx.[3, 8]

7. Given that dx = dx = 12, find a) dx b) dx + dx[24, 0]

Find the value of m for which dx = 0.[]

8. Given that dx = 5, evaluate a) dxb) dx.[9,  1]

Find the value of k for which dx = 31.[k = 3]

9. Given that dx = 10, find the value of k for which dx = 0.[k = ]

CATHOLICJUNIOR COLLEGE

ADDITIONAL MATHEMATICS 4018

AO MATHEMATICS 8174

INTEGRATION  TRIGONOMETRIC FUNCTIONS

SUMMARY

As integration is the reverse of differentiation, we have the following results.

Since sin x = cos xthen,  cos x dx = sin x + C

cos x =  sin x sin x dx =  cos x + C

tan x = sec2x sec2x dx = tan x + C

sec x = sec x tan x sec x tan x dx = sec x + C

cosec x =  cosec x cot x cosec x cot x dx =  cosec x + C

cot x =  cosec2x cosec2x dx =  cot x + C

In general,  f ‘ (x) sin [f(x)] dx =  cos [f(x)] + C

 f ‘ (x) cos[f(x)] dx = sin [f(x)] + C

 sec x dx = ln sec x + tan x + C.

ASSIGNMENT

1. Integrate with respect to x:

a) sin x + 2[ cos x + 2x + C]

b) 1  3 cos x[x 3 sin x + C]

c) cos x sin x[sin x + cos x + C]

d) sec2x 4 sin x[tan x + 4 cos x + C]

e) 3 cos x 2 sin x[3 sin x + 2 cos x + C]

f) 4 cos x + 3 sec2x[4 sin x + 3 tan x + C]

g) cos 2x[sin 2x + C]

h) sin 3x[cos 3x + C]

i) 2 cos 4x[sin 4x + C]

j) cos x[2 sin x + C]

k) sin x[ 2 cos x + C]

l) 2 cos (1 x)[ 2 sin (1 x) + C]

m) cos (1  2x)[sin (1  2x) + C]

n) cos (2x + )[sin (2x + ) + C]

o) 4 sin (x)[ 4 cos (x) + C]

2. Evaluate the following definite integrals:

a) dx[]

b) dx[ 2]

c) dx[]

3. Integrate with respect to x:

a) 2 sin2x[xsin 2x + C]

b) 4 cos2x[2x + sin 2x + C]

c) cos2 2x[x + sin 4x + C]

d) 6 sin2 2x[3xsin 4x + C]

e) sin2 3x[xsin 6x + C]

f) 2 cos2x[x + sin x + C]

g) (1 + 2 cos x)2[3x + 4 sin x + sin 2x + C]

h) (1  sin 2x)2 [x + cos 2xsin 4x + C]

i) 4 sin x cos x[ cos 2x + C]

j) sin 2x cos 2x[cos 4x + C]

k) (cos x + sin x)2[xcos 2x + C]

l) (cos x 2sin x) sin x[cos 2x+sin2xx+C]

4. Evaluate dx[1]

5. Show that dx = + 1.

CATHOLICJUNIOR COLLEGE

ADDITIONAL MATHEMATICS 4018

AO MATHEMATICS 8174

INTEGRATION  EXPONENTIAL AND LOGARITHM FUNCTIONS

SUMMARY

ASSIGNMENT

1. Integrate with respect to x:

a) ex + 1[ex + x + C]b) e2x[e2x + C]

c) 2e3x[e3x + C]d) ex ex[ ex ex + C]

e) e2x[ e2x + C]f) 2e1/2 x[4e1/2 x + C]

g) e2x + 1[e2x + 1 + C]h) 3e1 x[3e1 x + C]

i) 4e ½ (1 x)[8e ½ (1 x) + C]j) dx[e2 1]

k) dx[(e2 1)]l) dx[2(1 )]

m) dx[1 ]n) dx[]

o) dx[2]

2. Find y as a function of x given that = 1  3ex and that y = 4 when x = 0.[y = x 3ex + 7]

3. Find y as a function of x given that = e2x and that y = 6 when x = ln 3.[y = (e2x+ 3)]

4. Find esin x and hence find dx.[esin x+ C]

5. Find ex and hence find dx.[ex + C]

6. Integrate with respect to x:

a) [2 ln x + C]b) [ln (x + 1) + C]

c) [ln (2x 1) + C]d) [ln (2x + 1) + C]

e) [2 ln (3x + 2) + C]f) [ ln (1 x) + C]

7. Evaluate the following definite integrals:

a) dx[8 ln 2]b) dx[4  3 ln 2]

c) dx[ln 3]

8. Find ln(x2 + 1) and hence, evaluate dx.[ln 2]

9. Find ln(cos x) and hence, evaluate dx.[ln 2]

10. Find ln(ex + 1) and hence, evaluate dx.[ln ]

Miscellaneous Problems

11. <AJC 02/1/8> Let y = 2x2 ln x x2. Show that = 4x ln x. Hence, or otherwise,

show that x ln x dx = 2 ln 2 .

12. <NJC 02/1/11> Evaluate i) (6x 3)dxii) dx[31, ln ]

13. <AJC 02/1/11> Evaluate i) dxii) [sin  cos 3x] dx. [6, 1]

14. <JJC 02/2/5b> Evaluate dx.[]

15. <AJC 02/1/13b> Find the values of m if (3  2x) dx =  4x dx. [m = 4 or  1]

16. <JJC 02/1/9> Find i) dx ii) dx. [x5/2 + x1/2 + C, ln 2]

17. <NYJC 02/1/13> Evaluate i)  dx ii) dx. [1, ln ]

18. <PJC 02/1/13> Evaluate i) sin ( + )dx ii) dx iii) dx.

[ 2, 2, +  1]

19. <NJC 02/1/17> Find the coordinates of the stationary points of the curve y = .

Given that y = x, show that = . Hence, or otherwise, evaluate dx. [(0, 0), (4, 8), 36]

20. <RJC 02/1/18b> Given that y = x, show that = .

Hence evaluate dx.[]