Question

$$\underset{0}{\overset{2a}{\int }}f\left(x\right)dx$$  is equal to

Options

• $$2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$$

•  0

• $$\underset{0}{\overset{a}{\int }}f\left(x\right)dx+\underset{0}{\overset{a}{\int }}f(2a-x)dx$$

• $$\underset{0}{\overset{a}{\int }}f\left(x\right)dx+\underset{0}{\overset{2a}{\int }}f(2a-x)dx$$

Answer 1

$$\underset{0}{\overset{a}{\int }}f\left(x\right)dx+\underset{0}{\overset{a}{\int }}f(2a-x)dx$$

$$\text{According to the additivity property of integrals},$$
$${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx,wherea<c<b$$
using this property
$${\int }_0^{2a}f(x)dx={\int }_0^{a}f(x)dx+{\int }_0^{2a}f(x)dx......(1)$$
$$\text{Now, consider the integral},{\int }_0^{2a}f(x)dx$$
$$\text{Let }x=2a-t.Then,dx=d(2a-t)\Rightarrow dx=-dt$$
$$\text{Also, }x=a\Rightarrow t=aandx=2a\Rightarrow t=0$$
$$\text{Therefore, }{\int }_a^{2a}f(x)dx=-{\int }_a^{0}f(2a-t)dt$$
$$\Rightarrow {\int }_a^{2a}f(x)dx={\int }_0^{a}f(2a-t)dt$$
$$\Rightarrow {\int }_a^{2a}f(x)dx={\int }_0^{a}f(2a-x)dx$$
$$\text{Substituting this in equation (1) we get},$$
$${\int }_0^{2a}f(x)dx={\int }_0^{a}f(x)dx+{\int }_0^{a}f(2a-x)dx$$