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Question
`int_0^(2a)f(x)dx`
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Solution
We have.
`I=int_0^(2a)f(x)dx`
Then,
`I=int_0^af(x)dx+int_a^(2a)f(x)dx`
`I=int_0^af(x)dx+I_1 ......................["where, "I_1=int_a^(2a)f(x)dx]`
Let 2a - t = x then dx = - dt
If t = a ⇒ x = a
If t = 2a ⇒ x = 0
`I_1=int_0^(2a)f(x)dx=int_a^0f(2a-t)(-dt)=-int_a^0f(2a-t)dt`
`I_1=int_0^af(2a-t)dt=int_0^af(2a-x)dx`
`thereforeI=int_0^af(x)dx+int_0^af(2a-x)dx`
`I=int_0^af(x)dx+int_0^af(x)dx=2int_0^af(x)dx............................[f(2a-x=f(x))]`
Hence proved.
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