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# Choose the correct answers if $f(a+b-x) = f(x)$ , then $\int\limits_a^bx\;f(x)dx$ is equal to

$\begin{array}{1 1} (A)\frac{a+b}{2}\int\limits_a^bf(b-x)\;dx \\ (B)\frac{a+b}{2}\int\limits_a^bf(b+x)\;dx \\ (C)\frac{b-a}{2}\int\limits_a^bf(x)\;dx \\ (D)\frac{a+b}{2}\int\limits_a^bf(x)\;dx\end{array}$

Can you answer this question?

Toolbox:
• $(i) \int \limits_a ^ bf(x)dx=F(b)-F(a)$
• $(ii) \int \limits_a ^ bf(x)dx=\int \limits_a^b f(a+b-x)$
Given $I=\int \limits_a^b xf(x)dx$----------(1)

By applying the property $\int \limits_a^b f(a+b-x) =\int _a^b f(x) dx$

$I=\int \limits_a^b (a+b-x) f(a+b-x) dx$

$=\int \limits_a^b (a+b-x) f(x) dx \qquad \int \limits_a^b f(a+b-x)dx=\int \limits_a^b f(x) dx$----------(2)

Adding equ (1) and equ(2)

$2I=(a+b) \int \limits_a^b f(x)dx$

Therefore $I=\frac{a+b}{2} \int \limits_a ^b f(x)dx$

Hence the correct answer is D

answered Mar 6, 2013 by