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If f(a+b+1-x)=f(x), for all x, where a and b are fixed positive real numbers, then $\frac{1}{a+b}\int_{a}^{b}x(f(x)+f(x+1))dx$ is equal to :
Option: 1 $\int_{a-1}^{b-1}f(x)dx$

Option: 2 $\int_{a+1}^{b+1}f(x+1)dx$

Option: 3 $\int_{a-1}^{b-1}f(x+1)dx$

Option: 4 $\int_{a+1}^{b+1}f(x)dx$

Answers (1)

best_answer

Properties of the Definite Integral (Part 2) - King's Property -

Property 4 (King's Property)

This is one of the most important properties of definite integration.

$\\\mathbf{\int_{a}^{b}f(x)\;dx=\int_{a}^{b}f(a+b-x)\;dx}$

-

$f(a+b+1-x)=f(x)$

$\text{put x =1+x}$

$f(a+b-x)=f(x+1)$

$I=\frac{1}{a+b}\int_{a}^{b}\left [x(f(x)+f(x+1)) \right ]dx$ ...........1

$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b-x)(f(a+b-x)+f(a+b-x+1)) \right ]dx$

$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b-x)(f(x+1)+f(x)) \right ]dx$ ..........2

Adding 1 and 2

$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b)(f(x+1)+f(x)) \right ]dx$

$I=\int_{a}^{b}\left [(f(x+1)+f(x)) \right ]dx$

$\begin{array}{l}{2 \mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}+1) \mathrm{d} \mathrm{x}+\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}} \\ 2I={\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}+1-\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}} \\ {2 \mathrm{I}=2 \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}}\end{array}$

Correct Option (4)

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Ritika Jonwal

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