Question

If $f(a+b+1-x)=f\left(x\right)\forall x$, where a and b are fixed positive real numbers, then

$\frac{1}{a+b}{\int }_a^{b}x\left(f\right(x)+f(x+1\left)\right)dx$ is equal to.

• A. ${\int }_{a-1}^{b-1}f\left(x\right)dx$
• B. ${\int }_{a+1}^{b+1}f(x+1)dx$
• C. ${\int }_{a-1}^{b-1}f(x+1)dx$
• D. ${\int }_{a+1}^{b+1}f\left(x\right)dx$

Answer 1

The correct option is C

${\int }_{a-1}^{b-1}f(x+1)dx$

Explanation for correct Answer:

Step:1: Formation of Equation

$$\begin{gathered} f(a+b+1-x)=f\left(x\right) \\ putx=1+x \\ f(a+b-x)=f(x+1) \\ I=\frac{1}{a+b}{\int }_a^b\left[x\right(f\left(x\right)+f(x+1)\left)\right]dx..................\left(i\right) \\ I=\frac{1}{a+b}{\int }_a^b\left[\right(a+b-x\left)\right(f(a+b-x))+f(a+b-x+1)dx] \\ I=\frac{1}{a+b}{\int }_a^b\left[\right(a+b-x\left)\right(f(x+1)+f\left(x\right)\left)\right]...........\left(ii\right) \end{gathered}$$

Step:2:Adding the formed equation

Adding (i) and (ii),

$\Rightarrow 2I={\int }_a^{b}f(a+b-x+1)dx+{\int }_a^{b}f\left(x\right)dx$

$\Rightarrow 2I=2{\int }_a^{b}f\left(x\right)dx$

$\Rightarrow I={\int }_a^{b}f\left(x\right)dx$

put, $x=t+1,dx=dt$

$\Rightarrow I={\int }_{a-1}^{b-1}f(t+1)dt$

$\Rightarrow I={\int }_{a-1}^{b-1}f(x+1)dt$

$\frac{\mathbf{1}}{\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{x}\mathbf{\left(}\mathit{f}\mathbf{\right(}\mathit{x}\mathbf{)}\mathbf{+}\mathit{f}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{\left)}\mathbf{\right)}\mathbf{d}\mathit{x}$ is equal to ${\mathbf{\int }}_{\mathbf{a}\mathbf{-}\mathbf{1}}^{\mathbf{b}\mathbf{-}\mathbf{1}}\mathit{f}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{d}\mathit{x}$.

Hence, Option ‘C’ is correct.