Question

What is the value of the definite integral \[\int\limits_{\dfrac{1}{n}}^{\dfrac{{an - 1}}{n}} {\dfrac{{\sqrt x }}{{\sqrt {a - x} + \sqrt x }}} dx\] ?
A. \[\dfrac{a}{2}\]
B. \[\dfrac{{na + 2}}{{2n}}\]
C. \[\dfrac{{na - 2}}{{2n}}\]
D. None of these

Answer 1

Hint:Here, a definite integral is given. First, simplify the integral by applying the integration rule \[\int\limits_a^b {f\left( x \right) dx} = \int\limits_a^b {f\left( {a + b - x} \right) dx} \]. Then, add both integrals and solve the integral by using the sum rule of the integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right] dx} = \int\limits_a^b {f\left( x \right) dx} + \int\limits_a^b {g\left( x \right) dx} \]. After that, solve the integral by using the integration formula \[\int\limits_a^b {n dx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\]. In the end, apply the limits and solve the equation to get the required answer.



Formula Used:Integration rule: \[\int\limits_a^b {f\left( x \right) dx} = \int\limits_a^b {f\left( {a + b - x} \right) dx} \]
Sum rule of the integration:\[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right] dx} = \int\limits_a^b {f\left( x \right) dx} + \int\limits_a^b {g\left( x \right) dx} \]
\[\int\limits_a^b {n dx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\]



Complete step by step solution:The given definite integral is \[\int\limits_{\dfrac{1}{n}}^{\dfrac{{an - 1}}{n}} {\dfrac{{\sqrt x }}{{\sqrt {a - x} + \sqrt x }}} dx\].
Let consider,
\[I = \int\limits_{\dfrac{1}{n}}^{\dfrac{{an - 1}}{n}} {\dfrac{{\sqrt x }}{{\sqrt {a - x} + \sqrt x }}} dx\]
Simplify the upper limit.
\[I = \int\limits_{\dfrac{1}{n}}^{a - \dfrac{1}{n}} {\dfrac{{\sqrt x }}{{\sqrt {a - x} + \sqrt x }}} dx\] \[.....\left( 1 \right)\]
Now apply the integration rule \[\int\limits_a^b {f\left( x \right) dx} = \int\limits_a^b {f\left( {a + b - x} \right) dx} \].
\[I = \int\limits_{\dfrac{1}{n}}^{a - \dfrac{1}{n}} {\dfrac{{\sqrt {\dfrac{1}{n} + a - \dfrac{1}{n} - x} }}{{\sqrt {a - \left( {\dfrac{1}{n} + a - \dfrac{1}{n} - x} \right)} + \sqrt {\dfrac{1}{n} + a - \dfrac{1}{n} - x} }}} dx\]
\[ \Rightarrow I = \int\limits_{\dfrac{1}{n}}^{a - \dfrac{1}{n}} {\dfrac{{\sqrt {a - x} }}{{\sqrt {a - \dfrac{1}{n} - a + \dfrac{1}{n} + x} + \sqrt {a - x} }}} dx\]
\[ \Rightarrow I = \int\limits_{\dfrac{1}{n}}^{a - \dfrac{1}{n}} {\dfrac{{\sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}} dx\] \[.....\left( 2 \right)\]
Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[I + I = \int\limits_{\dfrac{1}{n}}^{a - \dfrac{1}{n}} {\dfrac{{\sqrt x }}{{\sqrt {a - x} + \sqrt x }}} dx + \int\limits_{\dfrac{1}{n}}^{a - \dfrac{1}{n}} {\dfrac{{\sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}} dx\]
Apply the sum rule of the integration.
\[2I = \int\limits_{\dfrac{1}{n}}^{a - \dfrac{1}{n}} {\left[ {\dfrac{{\sqrt x }}{{\sqrt {a - x} + \sqrt x }} + \dfrac{{\sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}} \right]} dx\]
\[ \Rightarrow 2I = \int\limits_{\dfrac{1}{n}}^{a - \dfrac{1}{n}} {\left[ {\dfrac{{\sqrt x + \sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}} \right]} dx\]
\[ \Rightarrow 2I = \int\limits_{\dfrac{1}{n}}^{a - \dfrac{1}{n}} 1 dx\]
Solve the integral by using the integration formula \[\int\limits_a^b {n dx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\].
\[ \Rightarrow 2I = \left[ x \right]_{\dfrac{1}{n}}^{a - \dfrac{1}{n}}\]
Apply the upper and lower limits.
\[ \Rightarrow 2I = a - \dfrac{1}{n} - \dfrac{1}{n}\]
\[ \Rightarrow 2I = a - \dfrac{2}{n}\]
\[ \Rightarrow 2I = \dfrac{{na - 2}}{n}\]
\[ \Rightarrow I = \dfrac{{na - 2}}{{2n}}\]
Thus, \[\int\limits_{\dfrac{1}{n}}^{\dfrac{{an - 1}}{n}} {\dfrac{{\sqrt x }}{{\sqrt {a - x} + \sqrt x }}} dx = \dfrac{{na - 2}}{{2n}}\].



Option ‘C’ is correct



Note: Students often get confused and try to solve the integral by splitting the function. Because of that, the integral became more complicated and they get the wrong answer.