
What is the value of the definite integral \[I = \int\limits_0^1 {x\left| {x - \dfrac{1}{2}} \right|} dx\]?
A. \[\dfrac{1}{3}\]
B. \[\dfrac{1}{4}\]
C. \[\dfrac{1}{8}\]
D. None of these
Answer
163.2k+ views
Hint: Here, a definite integral is given. The function present in the integral is an absolute value function. First, apply the absolute function formula that is \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\]. Then, break the interval for the absolute value function. After that, solve the integrals. In the end, apply the limits and solve it to get the required answer.
Formula Used: Formula used:
Absolute value function: \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]
Complete step by step solution:The given definite integral is \[I = \int\limits_0^1 {x\left| {x - \dfrac{1}{2}} \right|} dx\].
Substitute \[x = x - \dfrac{1}{2}\] in the absolute value function \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\].
We get,
\[\left| {x - \dfrac{1}{2}} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {x - \dfrac{1}{2}} \right),}&{if x - \dfrac{1}{2} < 0}\\{x - \dfrac{1}{2},}&{if x - \dfrac{1}{2} \ge 0}\end{array}} \right.\]
\[ \Rightarrow \left| {x - \dfrac{1}{2}} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {x - \dfrac{1}{2}} \right),}&{if x < \dfrac{1}{2}}\\{x - \dfrac{1}{2},}&{if x \ge \dfrac{1}{2}}\end{array}} \right.\]
The interval of the integration is 0 to 1.
This implies that \[\left| {x - \dfrac{1}{2}} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {x - \dfrac{1}{2}} \right),}&{if 0 < x < \dfrac{1}{2}}\\{x - \dfrac{1}{2},}&{if \dfrac{1}{2} \le x < 1}\end{array}} \right.\].
Break the integration by using the above absolute value function.
\[I = \int\limits_0^{\dfrac{1}{2}} { - x\left( {x - \dfrac{1}{2}} \right)} dx + \int\limits_{\dfrac{1}{2}}^1 {x\left( {x - \dfrac{1}{2}} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{1}{2}} {x\left( {\dfrac{1}{2} - x} \right)} dx + \int\limits_{\dfrac{1}{2}}^1 {x\left( {x - \dfrac{1}{2}} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{1}{2}} {\left( {\dfrac{1}{2}x - {x^2}} \right)} dx + \int\limits_{\dfrac{1}{2}}^1 {\left( {{x^2} - \dfrac{1}{2}x} \right)} dx\]
Now solve the integrals by using the formula \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
\[ \Rightarrow I = \left[ {\dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2}} \right) - \dfrac{{{x^3}}}{3}} \right]_0^{\dfrac{1}{2}} + \left[ {\dfrac{{{x^3}}}{3} - \dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2}} \right)} \right]_{\dfrac{1}{2}}^1\]
\[ \Rightarrow I = \left[ {\dfrac{{{x^2}}}{4} - \dfrac{{{x^3}}}{3}} \right]_0^{\dfrac{1}{2}} + \left[ {\dfrac{{{x^3}}}{3} - \dfrac{{{x^2}}}{4}} \right]_{\dfrac{1}{2}}^1\]
Apply the upper and lower limits.
\[ \Rightarrow I = \left[ {\left( {\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^2}}}{4} - \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^3}}}{3}} \right) - \left( {\dfrac{{{0^2}}}{4} - \dfrac{{{0^3}}}{3}} \right)} \right] + \left[ {\left( {\dfrac{{{1^3}}}{3} - \dfrac{{{1^2}}}{4}} \right) - \left( {\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^3}}}{3} - \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^2}}}{4}} \right)} \right]\]
\[ \Rightarrow I = \left[ {\left( {\dfrac{{\left( {\dfrac{1}{4}} \right)}}{4} - \dfrac{{\left( {\dfrac{1}{8}} \right)}}{3}} \right) - \left( {0 - 0} \right)} \right] + \left[ {\left( {\dfrac{1}{3} - \dfrac{1}{4}} \right) - \left( {\dfrac{{\left( {\dfrac{1}{8}} \right)}}{3} - \dfrac{{\left( {\dfrac{1}{4}} \right)}}{4}} \right)} \right]\]
\[ \Rightarrow I = \left( {\dfrac{1}{{16}} - \dfrac{1}{{24}}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{4}} \right) - \left( {\dfrac{1}{{24}} - \dfrac{1}{{16}}} \right)\]
\[ \Rightarrow I = \dfrac{1}{{16}} - \dfrac{1}{{24}} + \dfrac{1}{3} - \dfrac{1}{4} - \dfrac{1}{{24}} + \dfrac{1}{{16}}\]
\[ \Rightarrow I = \dfrac{2}{{16}} - \dfrac{2}{{24}} + \dfrac{1}{3} - \dfrac{1}{4}\]
\[ \Rightarrow I = \dfrac{1}{8} - \dfrac{1}{{12}} + \dfrac{1}{3} - \dfrac{1}{4}\]
\[ \Rightarrow I = \dfrac{{1 \times 3}}{{8 \times 3}} - \dfrac{{1 \times 2}}{{12 \times 2}} + \dfrac{{1 \times 8}}{{3 \times 8}} - \dfrac{{1 \times 6}}{{4 \times 6}}\]
\[ \Rightarrow I = \dfrac{3}{{24}} - \dfrac{2}{{24}} + \dfrac{8}{{24}} - \dfrac{6}{{24}}\]
\[ \Rightarrow I = \dfrac{{3 - 2 + 8 - 6}}{{24}}\]
\[ \Rightarrow I = \dfrac{3}{{24}}\]
\[ \Rightarrow I = \dfrac{1}{8}\]
Thus, \[I = \int\limits_0^1 {x\left| {x - \dfrac{1}{2}} \right|} dx = \dfrac{1}{8}\].
Option ‘C’ is correct
Note: Students often do mistake to integrating \[\int\limits_a^b {{x^n}dx} \]. They apply the formula \[\int\limits_a^b {{x^n}dx = \left[ {{x^{n + 1}}} \right]} _a^b\] which is an incorrect formula. The correct formula is \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
Formula Used: Formula used:
Absolute value function: \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]
Complete step by step solution:The given definite integral is \[I = \int\limits_0^1 {x\left| {x - \dfrac{1}{2}} \right|} dx\].
Substitute \[x = x - \dfrac{1}{2}\] in the absolute value function \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\].
We get,
\[\left| {x - \dfrac{1}{2}} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {x - \dfrac{1}{2}} \right),}&{if x - \dfrac{1}{2} < 0}\\{x - \dfrac{1}{2},}&{if x - \dfrac{1}{2} \ge 0}\end{array}} \right.\]
\[ \Rightarrow \left| {x - \dfrac{1}{2}} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {x - \dfrac{1}{2}} \right),}&{if x < \dfrac{1}{2}}\\{x - \dfrac{1}{2},}&{if x \ge \dfrac{1}{2}}\end{array}} \right.\]
The interval of the integration is 0 to 1.
This implies that \[\left| {x - \dfrac{1}{2}} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {x - \dfrac{1}{2}} \right),}&{if 0 < x < \dfrac{1}{2}}\\{x - \dfrac{1}{2},}&{if \dfrac{1}{2} \le x < 1}\end{array}} \right.\].
Break the integration by using the above absolute value function.
\[I = \int\limits_0^{\dfrac{1}{2}} { - x\left( {x - \dfrac{1}{2}} \right)} dx + \int\limits_{\dfrac{1}{2}}^1 {x\left( {x - \dfrac{1}{2}} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{1}{2}} {x\left( {\dfrac{1}{2} - x} \right)} dx + \int\limits_{\dfrac{1}{2}}^1 {x\left( {x - \dfrac{1}{2}} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{1}{2}} {\left( {\dfrac{1}{2}x - {x^2}} \right)} dx + \int\limits_{\dfrac{1}{2}}^1 {\left( {{x^2} - \dfrac{1}{2}x} \right)} dx\]
Now solve the integrals by using the formula \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
\[ \Rightarrow I = \left[ {\dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2}} \right) - \dfrac{{{x^3}}}{3}} \right]_0^{\dfrac{1}{2}} + \left[ {\dfrac{{{x^3}}}{3} - \dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2}} \right)} \right]_{\dfrac{1}{2}}^1\]
\[ \Rightarrow I = \left[ {\dfrac{{{x^2}}}{4} - \dfrac{{{x^3}}}{3}} \right]_0^{\dfrac{1}{2}} + \left[ {\dfrac{{{x^3}}}{3} - \dfrac{{{x^2}}}{4}} \right]_{\dfrac{1}{2}}^1\]
Apply the upper and lower limits.
\[ \Rightarrow I = \left[ {\left( {\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^2}}}{4} - \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^3}}}{3}} \right) - \left( {\dfrac{{{0^2}}}{4} - \dfrac{{{0^3}}}{3}} \right)} \right] + \left[ {\left( {\dfrac{{{1^3}}}{3} - \dfrac{{{1^2}}}{4}} \right) - \left( {\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^3}}}{3} - \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^2}}}{4}} \right)} \right]\]
\[ \Rightarrow I = \left[ {\left( {\dfrac{{\left( {\dfrac{1}{4}} \right)}}{4} - \dfrac{{\left( {\dfrac{1}{8}} \right)}}{3}} \right) - \left( {0 - 0} \right)} \right] + \left[ {\left( {\dfrac{1}{3} - \dfrac{1}{4}} \right) - \left( {\dfrac{{\left( {\dfrac{1}{8}} \right)}}{3} - \dfrac{{\left( {\dfrac{1}{4}} \right)}}{4}} \right)} \right]\]
\[ \Rightarrow I = \left( {\dfrac{1}{{16}} - \dfrac{1}{{24}}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{4}} \right) - \left( {\dfrac{1}{{24}} - \dfrac{1}{{16}}} \right)\]
\[ \Rightarrow I = \dfrac{1}{{16}} - \dfrac{1}{{24}} + \dfrac{1}{3} - \dfrac{1}{4} - \dfrac{1}{{24}} + \dfrac{1}{{16}}\]
\[ \Rightarrow I = \dfrac{2}{{16}} - \dfrac{2}{{24}} + \dfrac{1}{3} - \dfrac{1}{4}\]
\[ \Rightarrow I = \dfrac{1}{8} - \dfrac{1}{{12}} + \dfrac{1}{3} - \dfrac{1}{4}\]
\[ \Rightarrow I = \dfrac{{1 \times 3}}{{8 \times 3}} - \dfrac{{1 \times 2}}{{12 \times 2}} + \dfrac{{1 \times 8}}{{3 \times 8}} - \dfrac{{1 \times 6}}{{4 \times 6}}\]
\[ \Rightarrow I = \dfrac{3}{{24}} - \dfrac{2}{{24}} + \dfrac{8}{{24}} - \dfrac{6}{{24}}\]
\[ \Rightarrow I = \dfrac{{3 - 2 + 8 - 6}}{{24}}\]
\[ \Rightarrow I = \dfrac{3}{{24}}\]
\[ \Rightarrow I = \dfrac{1}{8}\]
Thus, \[I = \int\limits_0^1 {x\left| {x - \dfrac{1}{2}} \right|} dx = \dfrac{1}{8}\].
Option ‘C’ is correct
Note: Students often do mistake to integrating \[\int\limits_a^b {{x^n}dx} \]. They apply the formula \[\int\limits_a^b {{x^n}dx = \left[ {{x^{n + 1}}} \right]} _a^b\] which is an incorrect formula. The correct formula is \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for 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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

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

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
