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Prove that ∫(x)dx=∫f(x)dx+∫f(2a−x)dx - Mathematics and Statistics

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Prove that :

`int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`

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Solution

Since ‘a’ lies between 0 and 2a,
we have

`int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx,  .......(byint_a^bf(x)dx=int_a^cf(x)dx+int_c^bf(x)dx)`

`=I_1+I_2`     ........................(say)

`I_2=int_a^(2a)f(x)dx`

Put x=2a-t

therefore dx=-dt

When x=a,2a-t=a

t=a

When x =2a, 2a- t = 2a

t= 0

`I_2=int_0^(2a)f(x)dx=int_a^0f(2a-t)(-dt)`

`=-int_a^0f(2a-t)dt=int_0^af(2a-t)dt      ...................... (By int_a^bf(x)dx=-int_b^af(x)dx)`

`=int_0^af(2a-x)dx    ..............(By int_a^bf(X)dx=int_a^bf(t)dt)`

`int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`

`=int_0^a[f(x)+f(2a-x)]dx`

Concept: Evaluation of Definite Integrals by Substitution
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