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. $f\left(b\right)-f\left(a\right)$
• B. $g\left(b\right)-g\left(a\right)$
• 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

Answer: (C)

$\frac{\left[f\right(b{)}^2]-[f\left(a\right){]}^2}{2}$

$\frac{d}{dx}f\left(x\right)=g\left(x\right)\Rightarrow f\left(x\right)=\int g\left(x\right)$...(1)
$I={\int }_a^{b}f\left(x\right)g\left(x\right)dx$
$={\left[f\right(x)\int g\left(x\right)]}_a^b-{\int }_a^b\frac{d}{dx}f\left(x\right)\int g\left(x\right)dx$
Substituting values from (1)
$I={\left[{f\left(x\right)}^2\right]}_a^b-{\int }_a^{b}g\left(x\right)f\left(x\right)dx\Rightarrow 2I={\left[{f\left(x\right)}^2\right]}_a^b\Rightarrow I=\frac{{\left(f\left(b\right)\right)}^2-{\left(f\left(a\right)\right)}^2}{2}$