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Question
Prove that: `int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`
Solution
`LHS=int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx.........(1)`
Substitute x = a + t in the second integral
dx=dt
When x = a, t = 0.
When x = 2a, t = a.
`thereforeint_a^(2a)f(x)dx=int_0^af(a+t)dt`
`=int_0^af(a+(a-t))dt (therefore int_0^af(x)dx=int_0^a f(a-x)dx)`
`=int_0^af(2a-t)dt`
`int_a^(2a)f(x)dx=int_0^af(2a-x)dx (therefore int_0^af(t)dt=int_0^af(x)dx)`
Using the above in (1), we get
`int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx`
`=int_0^af(x)dx+int_0^af(2a-x)dx=RHS ("Proved")`
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