Question

If $\frac{d}{dx}f\left(x\right)=g\left(x\right)$ for $a\le x\le b$, then $\stackrel{b}{\underset{a}{\int }}f\left(x\right)g\left(x\right)dx$ equals to:

• A. $f\left(2\right)-f\left(1\right)$
• B. $g\left(2\right)-g\left(1\right)$
• C. $\frac{\left[f\right(b){]}^2-[f\left(a\right){]}^2}{2}$
• D. $\frac{\left[g\right(b){]}^2-[g\left(a\right){]}^2}{2}$

Answer 1

The correct option is C $\frac{\left[f\right(b){]}^2-[f\left(a\right){]}^2}{2}$

Given $\frac{d}{dx}f\left(x\right)=g\left(x\right)$

To find ${\int }_a^{b}f\left(x\right)g\left(x\right)dx$

Sol : ${\int }_a^{b}f\left(x\right)g\left(x\right)dx$

Let $f\left(x\right)=t$

$\frac{d}{dx}f\left(x\right)=\frac{dt}{dx}$

$g\left(x\right)=\frac{dt}{dx}$

$g\left(x\right)dx=dt$

$\int f\left(x\right)g\left(x\right)dx=\int tdt=\frac{t^2}{2}=\frac{\left(f\right(x){)}^2}{2}$

evaluating above integral area the limits a to b ,

${\int }_a^{b}f\left(x\right)g\left(x\right)dx=\frac{f(x{)}^2}{2}{|}_a^b=\frac{f(b{)}^2-f(a{)}^2}{2}$

$\therefore {\int }_a^{b}f\left(x\right)g\left(x\right)dx=\frac{f^2\left(b\right)-f^2\left(a\right)}{2}$