Question

\[\int\limits_0^{2a} f\left( x \right) dx\]  is equal to

Options

• \[2 \int\limits_0^a f\left( x \right) dx\]

•  0

• \[\int\limits_0^a f\left( x \right) dx + \int\limits_0^a f\left( 2a - x \right) dx\]

• \[\int\limits_0^a f\left( x \right) dx + \int\limits_0^{2a} f\left( 2a - x \right) dx\]

Answer 1

\[\int\limits_0^a f\left( x \right) dx + \int\limits_0^a f\left( 2a - x \right) dx\]

\[\text{According to the additivity property of integrals}, \]
\[ \int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx, where\ a < c < b\]
using this property
\[ \int_0^{2a} f(x)dx = \int_0^a f(x)dx + \int_0^{2a} f(x)dx . . . . . . (1)\]
\[\text{Now, consider the integral}, \int_0^{2a} f(x)dx\]
\[\text{Let }x = 2a - t . Then, dx = d(2a - t) \Rightarrow dx = - dt\]
\[\text{Also, }x = a \Rightarrow t = a\ and\ x\ = 2a \Rightarrow t = 0\]
\[\text{Therefore, }\int_a^{2a} f(x)dx = - \int_a^0 f(2a - t)dt\]
\[ \Rightarrow \int_a^{2a} f(x)dx = \int_0^a f(2a - t)dt\]
\[ \Rightarrow \int_a^{2a} f(x)dx = \int_0^a f(2a - x)dx\]
\[\text{Substituting this in equation (1) we get}, \]
\[ \int_0^{2a} f(x)dx = \int_0^a f(x)dx + \int_0^a f(2a - x)dx\]