Question

If $\frac{d}{dx}\left\{\begin{array}{c}f\left(x\right)\end{array}\right\}=g\left(x\right)$, then ${\int }_a^{b}f\left(x\right)g\left(x\right)dx$ is equal to

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

Answer 1

Answer: (A)

$\frac{1}{2}\left[f^2\right(b)-f^2(a\left)\right]$

$f\left(x\right)=\int g\left(x\right)dx$
${\int }_a^{b}f\left(x\right)g\left(x\right)dx=\left(f\right(x).f(x){)}_a^b-{\int }_a^{b}g(x).f(x)dx$
$I=f^2\left(b\right)-f^2\left(a\right)$
$I=\frac{1}{2}\left(f^2\right(b)-f^2(a\left)\right)$