INTEGRATION
THIRD TERM E-LEARNING NOTE
SUBJECT: FURTHER MATHEMATICS CLASS: SS 2
SCHEME OF WORK
WEEK SEVEN
TOPIC: INTEGRATION
Integration: This is defined as anti- differentiation. Suppose, y = x3 + 2x, the first derivative is
3x2 + 2. (dy/dx = 3x2 + 2)
then the anti – derivative of 3x2 + 2 = x3 + 2x
Thus, integration is the reverse process of differentiation and denoted by the symbol ∫.
If dy/dx = xn , then ∫ dy/dx = xn+1 + C
n + 1 (n ≠ -1)
where c is the arbitrary constant.
INDEFINITE INTEGRAL CONCEPTS
General Concept
Example: Evaluate the following integrals:
1. ∫x2 dx 2. ∫x5/2 dx 3.∫ 4/ x5 dx 4.∫ √x8 5.∫ (7x4 + 2) dx 6.∫(x5 + 2x4 – x3 + 6) dx
Solution;
1. x3 + C 2.x5/2 + 1 = 2x7/2 + c 3. ∫ 4x-5 dx = 4x-5+1= – 4x – 4 + C
3 5/2 + 1 7 -5 + 1
4. ∫ x4 = x4+1 = x5 + C 5. 7x4 +1 + 2x0+1 =7x5 + 2x + C
4 + 1 5
6. x6 + 2x5 – x4+ 6x + C
6 5 4
NB: Integral of a constant is not zero but the variable in the question.
Evaluation: Evaluate the following integrals; 1. ∫ (12x3 – x6 +1/x2) dx 2. ∫x2(3x2 + 4x) dx
Trigonometric integral:
The trigonometric integrals can be summarized in the following table. Remember that this is the reverse process of differentiation.
F(x) ∫ f(x)dx
Sin x – cos x + c
Cos x sin x + c
Sec2x tan x + c
Cosec2x – cot x + c
Sec x tan x sec x + c
exex + c
1/x ln x + c
Example: Evaluate each of the following integrals.
1. ∫ sin x – 5 cos x)dx 2. ∫ (5sinx + 3x2)dx
Solution:
1. . ∫ sin x – 5 cos x )dx = ∫sin x dx – ∫5 cos x dx
= -cos x – 5 ( sin x ) + c
= – cos x – 5sin x + c
2.∫ (5sinx + 3x2 )dx = ∫5 sin x dx + ∫3×2 dx
= -5cos x + x3 + c
3. ∫ e2x dx = e2x/2 + c
Evaluation: Evaluate the integrals:
1. ∫ (3 cos x + 2 sin x) dx 2. ∫ e2×2 + 5x dx
INTEGRATION BY ALGEBRAIC SUBSTITUTION
Sometimes integral are not given in the standard form, such integral are then reduced to standard form format before evaluation by algebraic substitution.
Suppose, an integral is given in the form ∫ f(ax + b)n dx
Then, the algebraic substitution is to represent the function in the bracket by any letter.
Let u = (ax + b) du/dx = a, dx = du/ a
∫un dx = ∫ un du/a
= 1/a ∫ un du
Example:
Evaluate the following integrals.
1. ∫ (2x2 – 5x )4 dx 2. ∫ 7 dx 3. ∫ xcos 2x2 dx 4. ∫ x2 √(x3 + 5)dx
(5x – 4 )5
Solution:
1. .∫ (2x2 – 5x )4 dx let u = 2x2 – 5x , du/dx = 4x – 5 , dx = du/4x -5
. ∫ u 4 du/ 4x -5 = u5+ c
5(4x – 5)
= (2x2 – 5x )5+ c
20x – 25
2. ∫ 7 dx let u = ( 5x – 4) , du/dx = 5, dx = du/5
(5x – 4 )5
∫ 7 u -5 du/5 = 7u-4+ c
5 x – 4
= 7(5x – 4 )-4
-20
3. ∫ x cos 2x2 dx let u = 2x2, du/dx = 4x, dx = du/4x
then; ∫x cos 2x2 dx = ∫x cos u du/4x = 1 x ( sin u ) = 1 sin 2x2 + c
4 4
4. ∫ x2 √(x3 + 5)dx let u = x3 + 5, du/dx = 3x2, dx = du/3x2
∫ x2 √(x3 + 5)dx = ∫x2 u1/2 du/3x2 = 1 x u3/2 = 2 (x3 + 5)3/2+ c
3/2 x 3 9
Evaluation:
Evaluate the following integrals:
1. ∫(5x – 7 )7/2dx 2. ∫cos 9x dx 3. ∫ xcos 2x dx.
INTEGRATION BY PARTS
This technique is uniquely useful in evaluating integrals that are not in the standard form. Such integrals cannot be solved by algebraic substitution.
From the product rule of differentiation, it can be generalized thus;
∫ vdu = uv – ∫ udv.
Example: Evaluate the following integral by parts.
∫ 2x sin x dx 2. ∫ e2xcos 2x dx
solution:
1. ∫ 2x sin x dx , let v = 2x, dv/dx = 2, dv = 2dx
∫du = sin x dx
u = -cos x
∫ vdu = uv – ∫ udv.
∫ x2 sin x = – x2cos x – ∫- cos x x 2 dx
= – x2cosx + 2∫ cos x
= – x2cos x + 2 sin x + c
2. ∫ x2ex dx , let v = x2, dv/dx = 2x, dv = 2xdx
du = ex u = ex dx
∫ vdu = uv – ∫udv
∫ x2ex = ex x2 – ∫ ex 2xdx
= x2 ex– 2∫ex x dx
the integral part in the RHS will have to be evaluated using integration by parts;
thus, v = x, dv/dx = 1, dv = dx , du =ex, u = ex
∫ vdu = uv – ∫udv
∫ x ex = ex . x – ∫exdx
= ex.x – ex
finally, ∫x2ex = x2 ex – 2(ex.x – ex)
= x2ex – 2xex + 2ex + c
Evaluation:
Evaluate 1. ∫x2cos x dx 2. ∫x3 e-x dx
INTEGRATION BY PARTIAL FRACTION
Sometimes rational functions are not expressed in the proper standard form; such function can be evaluated by transforming them into standard form through partial fractions. The knowledge of partial fractions is needed here to evaluate the functions.
Example; Integrate each of the following with respect to x;
1 2x + 32 x + 8
(2x + 1) (x – 1) ( x2 + 3x + 2)
solution:
1. resolve into partial fraction; 2x + 3 = A + B
(2x + 1)(x – 1) (2x + 1) (x – 1)
2x + 3 = A(x-1) + B(2x + 1)
when x = 1, 2(1) + 3 = B(2 + 1)
5 = 3B, B= 5/3
when x = -1/2, 2(-1/2) + 3 = A(-1/2 – 1 )
2 = – 3/2 A A = – 4/ 3
; 2x + 3 = ∫ -4 + ∫ 5 dx
(2x + 1)(x – 1) 3( 2x + 1) 3(x – 1)
= -4 ln (2x + 1) + 5 ln (x – 1)
2 x 3 3
= – 2/3 ln (2x + 1) + 5/3 ln ( x – 1) + c
2.x + 8 = x + 8 = A + B
(x2 + 3x + 2) (x + 1)(x + 2) x + 1 x + 2
x + 8 = A(x+2) + B(x+1)
when x = -2,
– 2 + 8 = B(-2+1)
6 = -B, B = – 6
when, x = -1, – 1 + 8 = A(-1 + 2)
7 = A.
thus, x + 8 =∫ 7 + ∫ – 6
x + 1 x + 2
= 7ln (x+ 1) – 6 ln(x+2) + c
Evaluation
Integrate by partial fraction.
1. 4x + 3 2. 1
(x – 3)(x+2) (x2+ 3x + 2)
GENERAL EVALUATION/REVISIONAL QUESTIONS
1.Find the derivatives of the following with respect to x;
(a) y = (15 + 5x)(1 + 2x) (b) y = (1 + 2x)12 (c) y = 3x2 (3 – 2x + 4x2) ½
2. Given that the gradient function of a curve is 8x – 2, find the equation of the curve at point (2, 4)
3. Find ∫(x2 + 1)(x3 – 2)dx
4. Find ∫x2 e2xdx
Reading Assignment: Read Integration, Page 31 – 46 Further Mathematics project III.
WEEKEND ASSINGMENT
1. Evaluate ∫ (x5 + 3)dx . A. x6/6 + 3x + c B. x5/6 + c C. x6/6 + c
2. Evaluate ∫ cos 7x dx A. 7sin 7x B. 1/7 sin 7x + c C. 7sin 7x + c
3. Integrate the function; (3x + 5)5wrt x .A 12(2x+3)6 + c B. (2x +3)6+ c C. (3x + 5)6 + c
12 18
4. Find ∫(x+1)(x2– 2)dx A. x4 + x3 – x2 – 2x + c B. x4 – x3 + x2 + c C. x + x4 – x3 – x2 + c
4 3 3 4
5. Integrate 1/x5 wrt x. A. x 6 + c B .x-4 + c C. x-4+ c
6 – 4 5
Theory:
1. Find ∫x sin2xdx 2. Evaluate ∫ dx
x ( x + 2)