Question
If $(\text{a}+\text{b}-\text{x})=\text{f}(\text{x}),$ then $\int\limits^\text{b}_\text{a}\text{x f}(\text{x})\text{dx}$ is equal to:

  1. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{b}-\text{x})\text{dx}$

  2. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{b}+\text{x})\text{dx}$

  3. $\frac{\text{b}-\text{a}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$

  4. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$

Answer

  1. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$

Solution:

Let, $\text{I}=\int\limits^\text{b}_\text{a}\text{x}\text{ f}(\text{x})\text{dx}\ ....(\text{i})$

$=\int\limits^\text{b}_\text{a}(\text{a}+\text{b}-\text{x})\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$

$=\int\limits^\text{b}_\text{a}(\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}\ ....(\text{ii})$

Adding (i) and (ii)

$2\text{I}=\int\limits^\text{b}_\text{a}(\text{x}+\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}$

$=(\text{a}+\text{b})\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$

Hence $\text{I}=\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$

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