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Question

Calculate 10x21x dx using reduction formulae.

A
4105
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B
4105
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C
16105
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D
16105
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Solution

The correct option is D 16105
We have to find 10x21xdx using reduction formula.

Let In=10xn1xdx

Take u=xn,dv=1xdx

du=nxn1dx,v=23(1x)3/2

In=[23xn(1x)3/2]10+2n310{(1x)3/2xn1}dx

=[00]+2n310{(1x)xn11x}dx

=2n310{(xn1xn)1x}dx

=2n310{xn11xxn1x}dx

In=2n310{xn11x}dx2n310{xn1x}dx

In=2n3In12n3In

3In=2nIn12nIn

(3+2n)In=2nIn1

In=2n3+2nIn1...(1)

We have to find I2=10x21xdx

Let us find I0=10x01xdx=101xdx=[23(1x)3/2]10=23

I1=10x1xdx

By (1), I1=25I0=25×23=415

I2=10x21xdx

By (1), I2=47I1=47×415=16105

Thus 10x21xdx=16105

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