
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
161.4k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025 Notes

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
