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Question

Evaluate : 22x21+5xdx.

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Solution

Let
I=22x21+5xdx .........(1)

By using property,

baf(x)dx=baf(a+bx)dx, we will get

I=22(2+2x)21+5(2+2x)dx

I=22x21+5xdx=22x21+15xdx

I=225x.x25x+1dx .................(2)

Adding (1) and (2), we will get,
2I=22[x21+5x+5x.x21+5x]dx

2I=22x2(1+5x)dx(1+5x)

2I=22x2 dx=220x2dx

[f(x)=x2 is an even function]

I=12×2[x33]20=12×2×(2)33
Hence,
I=83

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