Question

If f is a continuous function then, ${\int }_0^{2a}f\left(x\right)dx={\int }_0^{a}f\left(x\right)dx+{\int }_0^{2a}f(2a-x)dx.$

• A. True
• B. False

Answer 1

The correct option is B False

Let us try to see using the properties of definite integrals whether we are able to reach at the given equation.
$LHS={\int }_0^{2a}f\left(x\right)dx$
$={\int }_0^{a}f\left(x\right)dx+{\int }_a^{2a}f\left(x\right)dx$ (using property) ….(1)
Since it’s (2a - x) instead of (x) on the RHS we will try to make a substitution,
x = 2a - t
i.e., t = 2a - x
dx = -dt
Lower limit = 2a - a
= a.
Upper limit = 2a - 2a
= 0
$\therefore {\int }_a^{2a}f\left(x\right)dx={\int }_a^{0}f(2a-t)(-dt)$
$={\int }_0^{a}f(2a-t)dt$ (using property)
$={\int }_0^{a}f(2a-x)dx$ (using property) (2)
From (1) and (2)
LHS= ${\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f(2a-x)dx$
$\ne$ RHS
$\therefore$ The given statement is false.