Question

If $f$ is a function with period $P$, then ${\int }_a^{a+p}f\left(x\right)dx$ is

• A. equal to $f\left(a\right)$
• B. equal to $f\left(p\right)$
• C. independent of $a$
• D. None of these

Answer 1

Answer: (D)

None of these

f is a function with period p.

$\therefore f\left(x\right)=f(x-p)=f(x+p){\int }_a^{a+p}f\left(x\right)dx={\int }_a^{p}f\left(x\right)dx+{\int }_p^{a+p}f\left(x\right)dx.$

Now, $in{t}_p^{a+p}f\left(x\right)dx$

Let $x=p+u\Rightarrow dx=du$

$={\int }_0^{a}f(u+p)du={\int }_0^{a}f(x+p)dx\left[{\int }_a^{p}ft\right)dt={\int }_a^{b}f\left(x\right)dx]$

$\therefore {\int }_a^{a+b}f\left(x\right)dx={\int }_a^{p}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx[\because f(x+p)=f(x\left)\right]={\int }_0^{p}f\left(x\right)dx+{\int }_a^{p}f\left(x\right)dx={\int }_o^{p}f\left(x\right)dx$

which is independent of 'a'.