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Question

The value of π2π2x2cosx1+exdx is equal to

A
π242
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B
π24+2
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C
π2eπ2
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D
π2+eπ2
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Solution

The correct option is A π242
I=π2π2x2cosx1+exdx
I=π20(x2cosx1+ex+x2cosx1+ex)dx
I=π20(x2cosx+exx2cosx1+ex)dx
I=π20x2cosx dx
I=[x2sinx]π20π202xsinx dx
I=[x2sinx]π202[xcosx]π20π20(cosx) dx
I=π242[(00)+[sinx]π20]
I=π242

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