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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 

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)

Posted by

Ritika Jonwal

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