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**INTEGRAL CALCULUSINTEGRATION**

Integration is the reverse process of differentiation.

We know

Integration is the inverse operation of differentiation and is denoted by the symbol .

Hence, from equation (1), it follows that

i.e. Integral of xn with respect to variable x is equal to

Thus if we differentiate we can get back x^{n}.

Again if we differentiate and c being a constant, we get back the same x^{n}

Hence and this c is called the constant of integration

Integral calculus was primarily invented to determine the area bounded by the curves dividing the entire area into infinite number of infinitesimal small areas and taking the sum of all these small areas.**BASIC FORMULAS****Note:** In the answer for all integral sums we add +c (constant of integration) since the differentiation of constant is always zero.

**Elementary Rules:****Examples : **

Find**Solution:**

where c is arbitrary constant

It is sometime possible by a change of independent variable to transform a function into another which can be readily integrated.

We can show the following rules.

To put z = f (x) and also adjust dz = f'(x) dx

**Example:** ∫F{ h(x )} h'(x ) dx, take ez = h(x) and to adjust dz = h'(x) dx

then integrate F(z) d using normal rule.**Example: **

We put (2x + 3) = t ⇒ so 2 dx = dt or dx = dt / 2

Therefore

This method is known as Method of Substitution**Example:**

We put (x^{2} +1) = t

so 2x dx = dt or x dx = dt / 2**IMPORTANT STANDARD FORMULAE****Examples:**** where z=e ^{x} dz = e^{x} dx**

where u and v are two different functions of x

[Here degree of the numerator must be lower than that of the denominator; the denominator contains non–repeated linear factor]

so 3x + 2 = A (x – 3) + B (x – 2)

We put x = 2 and get

3.2 + 2 = A (2–3) + B (2–2) => A = –8

we put x = 3 and get

3.3 +2 = A (3–3) + B (3–2) => B= 11

or 3x + 2 = A (x – 2) (x – 3) + B (x – 3) +C (x – 2)^{2}

Comparing coefficients of x2, x and the constant terms of both sides, we find

A+C = 0 …………(i)

–5A + B – 4C = 3 ……(ii)

6A – 3B + 4C = 2 …….(iii)

By (ii) + (iii) A – 2B = 5 ..…….(iv)

(i) – (iv) 2B + C = –5 …….(v)

From (iv) A = 5 + 2B

From (v) C = –5 – 2B

From (ii) –5 ( 5 + 2B) + B – 4 (– 5 – 2B) = 3

or – 25 – 10B + B + 20 + 8B = 3

or – B – 5 = 3

or B = – 8, A = 5 – 16 = – 11, from (iv) C = – A = 11**Type III:**

so 3x^{2} –2x +5 = A (x^{2} + 5 ) + (Bx +C) (x–1)

Equating the coefficients of x^{2}, x and the constant terms from both sides we get

A + B = 3 …………(i)

C – B = –2 …………(ii)

5A – C = 5 ………….(iii)

by (i) + (ii) A + C = 1 ……… (iv)

by (iii) + (iv) 6A = 6 ……… (v)

or A = 1

therefore B = 3 – 1 = 2 and C = 0**Example:****Solution:**** **we put x^{3} = z, 3x^{2} dx = dz**Example:** Find the equation of the curve where slope at (x, y) is 9x and which passes through the origin.**Solution:**

Since it passes through the origin, c = 0; thus required eqn . is 9x^{2} = 2y.**DEFINITE INTEGRATION**

Suppose F(x) dx = f (x)

As x changes from a to b the value of the integral changes from f (a) to f (b). This is as

‘b’ is called the upper limit and ‘a’ the lower limit of integration. We shall first deal with

indefinite integral and then take up definite integral.**Example:****Solution: ****Note: In definite integration the constant (c) should not be added****Example:****Solution: **

Now,**IMPORTANT PROPERTIES****Important Properties of definite Integral****Example:****Solution: ****Example: **Evaluate **Solution:**

let x^{5} = t so that 5x^{4} dx = dt

(by standard formula b)

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