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# BernoulliŌĆÖs Formula

BernoulliŌĆÖs formula for integration of product of two functions is below

BernoulliŌĆÖs Formula

The evaluation of an indefinite integral of the form Ōł½u (x) v ( x)dx becomes very simple, when u is a polynomial function of x (that is, u ( x) = a0 xn + a1 xnŌłÆ1 + + an ) and v ( x) can be easily integrated successively. It is accomplished by a formula called BernoulliŌĆÖs formula. This formula is actually an extension of the formula of integration by parts. To derive the formula, we use the following notation: Proceeding in this way, we get

Ōł½ uvdx = uv(1) ŌłÆ u (1)v( 2) + u ( 2)v( 3) ŌłÆ u (3)v( 4) + .

The above result is called the BernoulliŌĆÖs formula for integration of product of two functions.

Note

Since u is a polynomial function of x , the successive derivative u( m) will be zero for some positive integer m and so all further derivatives will be zero only. Hence the right-hand-side of the above formula contains a finite number of terms only.

Example 9.31

Evaluate ŽĆŌł½0 x2 cos nx dx , where n is a positive integer.

Solution

Taking u = x2 and v = cos nx , and applying the BernoulliŌĆÖs formula, we get ### Example 9.32

Evaluate : 1Ōł½0 eŌłÆ2x (1 + x ŌłÆ 2x3 ) dx .

### Solution

Taking u = 1 + x ŌłÆ 2x3 and v = eŌłÆ2x , and applying the BernoulliŌĆÖs formula, we get

I = 1Ōł½0 eŌłÆ2x (1+ x ŌłÆ 2x3 ) dx Example 9.33

Evaluate : 2ŽĆŌł½0 x2 sin nx dx , where n is a positive integer.

Solution

Taking u = x2 and v = sin nx , and applying the BernoulliŌĆÖs formula, we get Example 9.34

Evaluate : 1Ōł½ŌłÆ1 eŌłÆ ╬╗x (1 ŌłÆ x2 ) dx .

### Solution

Taking u = 1ŌłÆ x2 and v = eŌłÆ╬╗ x , and applying the BernoulliŌĆÖs formula, we get Tags : Applications of Integration , 12th Maths : UNIT 9 : Applications of Integration
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12th Maths : UNIT 9 : Applications of Integration : BernoulliŌĆÖs Formula | Applications of Integration