Question

If $y=f\left(x\right)$ and $y=g\left(x\right)$ are symmetrical about the line $x=\frac{\alpha +\beta }{2},$ then $\underset{\alpha }{\overset{\beta }{\int }}f\left(x\right)g^{\prime }\left(x\right)dx$ is equal to

• A. $\underset{\alpha }{\overset{\beta }{\int }}{f}^{\prime }\left(x\right)g\left(x\right)dx$
• B. $-\underset{\alpha }{\overset{\beta }{\int }}{f}^{\prime }\left(x\right)g\left(x\right)dx$
• C. $\frac{1}{2}\underset{\alpha }{\overset{\beta }{\int }}\left(f\right(x\left)g^{\prime }\right(x)-f^{\prime }(x\left)g\right(x\left)\right)dx$
• D. $\frac{1}{2}\underset{\alpha }{\overset{\beta }{\int }}\left(f\right(x\left)g^{\prime }\right(x)+f^{\prime }(x\left)g\right(x\left)\right)dx$

Answer 1

Answer: (A), (B), (C), (D)

$\frac{1}{2}\underset{\alpha }{\overset{\beta }{\int }}\left(f\right(x\left)g^{\prime }\right(x)+f^{\prime }(x\left)g\right(x\left)\right)dx$

$f\left(x\right)=f(\alpha +\beta -x)\Rightarrow f\left(\alpha \right)=f\left(\beta \right)$
and $g\left(x\right)=g(\alpha +\beta -x)\Rightarrow g\left(\alpha \right)=g\left(\beta \right)$

$I=\underset{\alpha }{\overset{\beta }{\int }}f\left(x\right)g^{\prime }\left(x\right)dx\dots \left(1\right)={f\left(x\right)g\left(x\right)|}_{\alpha }^{\beta }-\underset{\alpha }{\overset{\beta }{\int }}{f}^{\prime }\left(x\right)g\left(x\right)dx=\left[f\right(\beta \left)g\right(\beta )-f(\alpha \left)g\right(\alpha \left)\right]-\underset{\alpha }{\overset{\beta }{\int }}{f}^{\prime }\left(x\right)g\left(x\right)dx$
$\Rightarrow I=-\underset{\alpha }{\overset{\beta }{\int }}{f}^{\prime }\left(x\right)g\left(x\right)dx$
Also, since $f^{\prime }\left(x\right)=-f^{\prime }(\alpha +\beta -x)$
$I=\underset{\alpha }{\overset{\beta }{\int }}{f}^{\prime }\left(x\right)g\left(x\right)dx\dots \left(2\right)$

From $\left(1\right)+\left(2\right),$
$2I=\underset{\alpha }{\overset{\beta }{\int }}f\left(x\right)g^{\prime }\left(x\right)dx+\underset{\alpha }{\overset{\beta }{\int }}{f}^{\prime }\left(x\right)g\left(x\right)dx\Rightarrow I=\frac{1}{2}\underset{\alpha }{\overset{\beta }{\int }}\left(f\right(x\left)g^{\prime }\right(x)+f^{\prime }(x\left)g\right(x\left)\right)dx\Rightarrow I=\frac{1}{2}\underset{\alpha }{\overset{\beta }{\int }}\left(f\right(x\left)g^{\prime }\right(x)-f^{\prime }(x\left)g\right(x\left)\right)dx$
$\{\because f^{\prime }(x)=-f^{\prime }(\alpha +\beta -x\left)\right\}$