Question

If f is a continuous function in the interval [a, b], then the value of ${\int }_0^{1}f\left(\left(b-a\right)x+a\right)\mathrm{dx}$ is equal to

• A. $\left(b-a\right){\int }_a^{b}f\left(x\right)\mathrm{dx}$
• B. $\left(a-b\right){\int }_a^{b}f\left(x\right)\mathrm{dx}$
• C. $\frac{1}{\left(}{\int }_a^{b}f\left(x\right)\mathrm{dx}$
• D. $\frac{1}{\left(}{\int }_b^{a}f\left(x\right)\mathrm{dx}$

Answer 1

Answer: (C), (D)

The correct options are
C $\frac{1}{\left(}{\int }_a^{b}f\left(x\right)\mathrm{dx}$
D $\frac{1}{\left(}{\int }_b^{a}f\left(x\right)\mathrm{dx}$

We have,
${\int }_0^{1}f\left(\left(b-a\right)x+a\right)\mathrm{dx}\mathrm{Let}\left(b-a\right)x+a=t\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{b-a}.\mathrm{Also},\mathrm{when}x=0,t=a\mathrm{and},\mathrm{when}x=1,t=b.\therefore {\int }_0^{1}f\left(\left(b-a\right)x+a\right)\mathrm{dx}={\int }_a^{b}f\left(t\right)\frac{\mathrm{dt}}{b-a}=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)\mathrm{dx}=-\frac{1}{a-b}{\int }_a^{b}f\left(x\right)\mathrm{dx}=\frac{1}{a-b}{\int }_b^{a}f\left(x\right)\mathrm{dx}$