Question

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 1

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 \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].