INTEGRATION [INDEFINITE Integral DEFINITE INTEGRAL AND AREA UNDER CURVE]

THIRD TERM E-LEARNING NOTE

SUBJECT: FURTHER MATHEMATICS CLASS: SS 2

SCHEME OF WORK

WEEK EIGHT

TOPIC : INTEGRATION [INDEFINITE Integral DEFINITE INTEGRAL AND AREA UNDER CURVE]

The process of reversing differentiation is called Integration. If dy/dx = 3x², then y could be x³, as the derivative of x³is 3x².

We say that x³ is an integral of 3x² with respect to x. The symbol for integration sign is given by ∫ . The expression to be integrated is put between the ∫ sign and dx.

∫ 3×2 dx could be x³

Since differentiating any constant gives zero, the following also have derivation 3x².

X³ + 2, x³ + 4.5, x³ – 17etc

In general, any function of the form x3 + c, where c is the constant has derivative of 3x²

Hence, ∫ 3x² dx = x³ + C. C is called constant of integration. Because we do not know the actual or definite value of C, this is called INDEFINITE INTEGRAL.

Let y = xⁿ⁺¹/ n+1 , Differentiating dy/dx = (n+1) xⁿ⁺¹/n+1 = xⁿ

Reversing this,

∫ xⁿ dx = xⁿ⁺¹/n+1 + C

∫ kxn dx = K ^xn+1/n+1 + C

To Integrate a sum, integrate each term as this is similar to differentiating a sum.

Examples:

Evaluate (i) ∫(2x³ + 3x² – 4) dx

2/4x⁴ + 3/3x³ – 4x + C

½ x2 + x3 – 4x + C

Evaluate ∫(4t³ + 2t² + ½ t²) dt

4/4 t4 + 2/3 t3 + ½ × ½ t² + C

t⁴ + 2t3 + ¼ t2 + C.

Integration of basic Trigonometric functions

Consider the table below for differentiating trigonometric functions.

Consider a similar table as the one given above

Example

∫(cos2x + sin3x) dx

∫cos2x dx + ∫ sin3x dx

½ sin2x + -1/3 cos3x

½ sin2x – 1/3 cos3x + K.

Sometimes,the value of the constant C can be found, if extra information is given.

If dy/dx = x2+2x-3, find y in terms of x given that x = 1, y =4

Y =∫(x2 +2x-3)dx

Y =x3/3 + 2x/2 – 3x + C

Y = x3/3 +x2 -3x + C

Putting x=1,and y = 4

4 = 1/3 + 1 – 3 + C. hence; C = 5 2/

;. 1/3×3 – x2 – 3x + 5 2/3.

If dy/dx = 2Cos3x, find y given that y =2 and x = 1/6∏

∫2Cos3x = 2/3Sin3x + C

2 = 2/3Sin3(∏/6) + C = 2/3 Sin∏/2 + C

Multiplying by ½ ∏ (180/∏) = 90⁰

Sin ½ ∏ – Sin 90⁰ = 1

2 = 2/3 ×1 + C

C = 4/3

Y = 2/3Sin3x + 4/3 .

Evaluation:

Evaluate these indefinite integrals

∫(y2 – 7y)dy

∫(3×2 – 2x – 1)dx

∫(Cos4x)dx

∫(3cos2x + 4sin3x)dx

If dy/dx = 4×2 + 1 and y = 2 when x =3, find y in terms of x

If dy/dx = 2Sin1/3x and y = 4 when x =∏m, find y

DEFINITE INTEGRALS

In this part, the constant is removed. If a definite integration is performed, the function is evaluated between the values called limits. Upper and lower ie

Example: Evaluate

=x³ + C , (3³ + c) – (2³ + c)

(27 + C) – (8 + C)

27 + C -8 – C

19.

= 19.

= { ½ Sin2}

(½ sin× 0)

½ sin

= ½

Area under curve using Definite Integral

Given in the diagram, the area between the curve and the x axis from x = a and to x = b. The area is given by

Area =

The area can be explained as: Area = ∫ y × dx = Sum of height of rectangle × width of rectangle

y

a b

Ex 1: Find the Area between the curve y = x3 – x and the x axis when x =2 and x = 4.

y

0 2 4 x

{ ¼ (4)⁴ – ½(4)2} – { ¼ X 2⁴ – ½ (2)²}

64 – 8 -4 + 2 = 54 UNITS.

Ex 2 : Find the area in the diagram shown below

y = 4 – x²

4

-2 0 2

∫(4 – x2 )dx

{4x – x3/3}₂².

4(2) – 2³/3 – 4(-2) – (-2)³/3

(8- 8/3) – (-8 + 8/3)

8 – 8/3 + 8 – 8/3

32/3.square unitrs

Evaluation:

Find the area between the values as shown below

y

y = 3x²

0 2 4 x

Find the area between the curve y = x² + 3 At the xs axis and when x = -1 and x = 3.

Evaluation

1. x² and the line y = 2.

General Evaluation:

∫(3x – 1)(x + 2) dx

∫5cos4x (dx)

Reading Assignment :Solve the evaluation questions given above

Weekend Assignment:

Evaluate A. 2/3 B. -2/3 C. -6 2/3 D. 6 2/3

Evaluate A. 4 B. 2 C. 4/3 D. 1/3

Evaluate A. – ½ B. 1 C. -1 D. 0

Find the area enclosed by by the curve y = x² , X = 0 and X = 3 A. 9 B. 7 C. 5/2 D. 5

Given y = 3x -2, x=3, x=4. Find the area under the curve A. 4/3 B. 17/2 C. 6 D. 3

Theory

Find the area enclosed between the curve y =x2 + x -2 and the x axis

Find the area enclosed by the curve y = x2 – 3x + 3 and the y = 1.