Question

Prove that $\int\limits_0^a {f\left( x \right)dx} = \int\limits_0^a {f\left( {a - x} \right)dx}$ and hence evaluate $\int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx}$.

Answer 1

We begin by letting $x = a - t$ and the corresponding value of limits. Substitute these values in LHS of the given expression. Apply rules of integration and prove it equal to RHS. Then, let $I = \int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx}$. Replace $x$ by $x - a$ and find the value of corresponding integration. Add both the integrations to get a simplified expression. Apply the formula of integration and then put limits to get the required answer.

Complete step-by-step answer:
We have to prove $\int\limits_0^a {f\left( x \right)dx} = \int\limits_0^a {f\left( {a - x} \right)dx}$
Let $x = a - t$
Then, differentiate both sides,
$dx = - dt$
When $x = 0$, then the value of $t$ is
$0 = a - t \\\ \Rightarrow t = a \\\$
And when $x = a$, then the value of $t$ is
$a = a - t \\\ \Rightarrow t = 0 \\\$
Substituting the values in LHS of the expression.
$\int\limits_0^a {f\left( x \right)dx} = \int\limits_a^0 {f\left( {a - t} \right)\left( { - dt} \right)} = - \int\limits_a^0 {f\left( {a - t} \right)\left( {dt} \right)}$
We can interchange the limits by multiplying a negative sign.
$\int\limits_0^a {f\left( {a - t} \right)\left( {dt} \right)}$
On substituting back the value of $t = x$, we will get,
$\int\limits_0^a {f\left( {a - x} \right)\left( {dt} \right)}$, which is equal to RHS.

Now, we have to find the value of $\int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx}$
Let $I = \int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx}$ eqn. (1)
Replace $x$ by $x - a$ in the above equation, as $\int\limits_0^a {f\left( x \right)dx} = \int\limits_0^a {f\left( {a - x} \right)dx}$
$I = \int\limits_0^a {\dfrac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt {a - \left( {a - x} \right)} }}dx}$
$\Rightarrow I = \int\limits_0^a {\dfrac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}dx}$ eqn.(2)
Adding equation (1) and (2),
$2I = \int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx} + \int\limits_0^a {\dfrac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}dx} \\\ \Rightarrow 2I = \int\limits_0^a {\dfrac{{\sqrt x + \sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}dx} \\\ \Rightarrow 2I = \int\limits_0^a {dx} \\\$
Now, $\int {dx = x}$
$2I = \left[ x \right]_0^a$
We will put the limits and divide the equation by 2.
$2I = \left( {a - 0} \right) \\\ \Rightarrow 2I = a \\\ \Rightarrow I = \dfrac{a}{2} \\\$
Hence, the value of $\int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx}$ is $\dfrac{a}{2}$.

Note: Students must know the basic rules of integration, how to change limits and formulas of integration to do these types of questions. Also, we have to find the value of $\int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx}$ using the property proved in previous step. While applying the limits, we apply the upper limit first and then subtract the value of lower limit from it.