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Question

Evaluate the integral
98x89xdx

A
π
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B
π3
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C
π2
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D
π62
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Solution

The correct option is C π2

I=98x89xdx ---(1)

baf(x)dx=baf(a+bx)dx

I=989xx8dx ---(2)

Adding 1 and 2, we get

2I=98x8+9x(x8)(9x)

2I=981(x8)(9x)dx=98dx(0.5)2(x8.5)2

=sin1(x8.50.5)98

2I=sin1(1)sin1(1)

I=π2


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