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. ∫x^5/2 dx 3.∫ 4/ x5 dx 4.∫ √x⁸ 5.∫ (7x⁴ + 2) dx 6.∫(x⁵ + 2x⁴ – x³ + 6) dx

Solution;

1. x³ + C 2.x^{5/2 + 1} = 2x^7/2 + c 3. ∫ 4x^-5 dx = 4x^-5+1= – 4^{x – 4} + C

3 5/2 + 1 7 -5 + 1

4. ∫ x⁴ = x⁴⁺¹ = x⁵ + C 5. 7x^{4 +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

e^xe^x + 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. ∫ e^2x dx = e^2x/2 + c

Evaluation: Evaluate the integrals:

1. ∫ (3 cos x + 2 sin x) dx 2. ∫ e^{2×2 + 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 u^1/2 du/3x² = 1 x u^3/2 = 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. ∫ e^2xcos 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. ∫ x²e^x dx , let v = x², dv/dx = 2x, dv = 2xdx

du = e^x u = e^x dx

∫ vdu = uv – ∫udv

∫ x²e^x = e^x x² – ∫ e^x 2xdx

= x² e^x– 2∫e^x 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 =e^x, u = e^x

∫ vdu = uv – ∫udv

∫ x e^x = e^x . x – ∫e^xdx

= e^x.x – e^x

finally, ∫x²e^x = x² e^x – 2(e^x.x – e^x)

= x²e^x – 2xe^x + 2e^x + 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 e^2xdx

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)