Question

What is the value of the integral \[\int\limits_2^3 {\dfrac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} dx\]?
A. 1
B. 0
C. \[ - 1\]
D. \[\dfrac{1}{2}\]

Answer 1

Hint: Here, a definite integral is given. First, simplify the integral by using the integral rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\]. Then, add this integral with the given integral and simplify it. After that, solve the integral using the integration formulas. In the end, apply the upper and lower limits to get the value of the given integral.



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



Complete step by step solution:The given definite integral is \[\int\limits_2^3 {\dfrac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} dx\].
Let consider,
\[I = \int\limits_2^3 {\dfrac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} dx\] \[.....\left( 1 \right)\]
Apply the integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] on the right-hand side.
\[I = \int\limits_2^3 {\dfrac{{\sqrt {2 + 3 - x} }}{{\sqrt {5 - \left( {2 + 3 - x} \right)} + \sqrt {2 + 3 - x} }}} dx\]
\[ \Rightarrow I = \int\limits_2^3 {\dfrac{{\sqrt {5 - x} }}{{\sqrt x + \sqrt {5 - x} }}} dx\] \[.....\left( 2 \right)\]
Now add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[I + I = \int\limits_2^3 {\dfrac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} dx + \int\limits_2^3 {\dfrac{{\sqrt {5 - x} }}{{\sqrt x + \sqrt {5 - x} }}} dx\]
\[ \Rightarrow 2I = \int\limits_2^3 {\left[ {\dfrac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }} + \dfrac{{\sqrt {5 - x} }}{{\sqrt x + \sqrt {5 - x} }}} \right]} dx\]
\[ \Rightarrow 2I = \int\limits_2^3 {\left[ {\dfrac{{\sqrt x + \sqrt {5 - x} }}{{\sqrt {5 - x} + \sqrt x }}} \right]} dx\]
\[ \Rightarrow 2I = \int\limits_2^3 1 dx\]
Apply the integration formula \[\int\limits_a^b n dx = \left[ {nx} \right]_a^b\].
\[2I = \left[ x \right]_2^3\]
\[ \Rightarrow 2I = \left[ {3 - 2} \right]\]
\[ \Rightarrow 2I = 1\]
\[ \Rightarrow I = \dfrac{1}{2}\]
Therefore, \[\int\limits_2^3 {\dfrac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} dx = \dfrac{1}{2}\].



Option ‘D’ is correct



Note: Students often get confused about the formula of the definite integral of the function. They used \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) + F\left( a \right)\] , which is incorrect. The correct formula is \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
Sometimes they also add integration constant \[c\] in the definite integral. But definite integral is calculated for a certain interval. So, there is no need to write the integration constant.