Question

$\underset{0}{\overset{x}{\int }}\left\{\underset{0}{\overset{u}{\int }}f\right(t\left)dt\right\}du$ is equal to:

• A. $\underset{0}{\overset{x}{\int }}(x-u)f\left(u\right)du$
• B. $\underset{0}{\overset{x}{\int }}uf(x-u)du$
• C. $x\underset{0}{\overset{x}{\int }}f\left(u\right)du$
• D. $x\underset{0}{\overset{x}{\int }}uf(u-x)du$

Answer 1

Answer: (A), (B)

$\underset{0}{\overset{x}{\int }}uf(x-u)du$

$I=\underset{0}{\overset{x}{\int }}1\cdot \left\{{\int }_0^{u}f\right(t\left)dt\right\}du$
Integrating by parts choosing $1$ as the second function
$\Rightarrow I={\left\{u\underset{0}{\overset{u}{\int }}f\left(t\right)dt\right\}}_0^x-\underset{0}{\overset{x}{\int }}f\left(u\right)udu$
$=x\underset{0}{\overset{x}{\int }}f\left(t\right)dt-\underset{0}{\overset{x}{\int }}f\left(u\right)udu$
$=x\underset{0}{\overset{x}{\int }}f\left(u\right)du-\underset{0}{\overset{x}{\int }}f\left(u\right)udu\Rightarrow I=\underset{0}{\overset{x}{\int }}f\left(u\right)(x-u)du$
Using the property, $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$
$\therefore I=\underset{0}{\overset{x}{\int }}uf(x-u)du$