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

Hint: 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.