Question

What is the value of the definite integral is \[\int\limits_0^1 {\left| {3{x^2} - 1} \right|} dx\]?
A. 0
B. \[\dfrac{4}{{3\sqrt 3 }}\]
C. \[\dfrac{3}{7}\]
D. \[\dfrac{5}{6}\]

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 greatest integer function. After that, solve the integrals. In the end, apply the limits and solve it to get the required answer.



Formula Used:Absolute value function: \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]
\[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\]



Complete step by step solution:The given definite integral is \[\int\limits_0^1 {\left| {3{x^2} - 1} \right|} dx\].
Let consider,
\[I = \int\limits_0^1 {\left| {3{x^2} - 1} \right|} dx\]
Substitute \[x = 3{x^2} - 1\] 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,
\[\left| {3{x^2} - 1} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {3{x^2} - 1} \right),}&{if 3{x^2} - 1 < 0}\\{3{x^2} - 1,}&{if 3{x^2} - 1 \ge 0}\end{array}} \right.\]
\[ \Rightarrow \left| {3{x^2} - 1} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {3{x^2} - 1} \right),}&{if {x^2} < \dfrac{1}{3}}\\{3{x^2} - 1,}&{if {x^2} \ge \dfrac{1}{3}}\end{array}} \right.\]
\[ \Rightarrow \left| {3{x^2} - 1} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {3{x^2} - 1} \right),}&{if x < \dfrac{1}{{\sqrt 3 }}}\\{3{x^2} - 1,}&{if x \ge \dfrac{1}{{\sqrt 3 }}}\end{array}} \right.\]
The interval of the integration is 0 to 1.
This implies that \[\left| {3{x^2} - 1} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {3{x^2} - 1} \right),}&{if 0 < x < \dfrac{1}{{\sqrt 3 }}}\\{3{x^2} - 1,}&{if \dfrac{1}{{\sqrt 3 }} < x \le 1}\end{array}} \right.\].
Break the integration by using the above absolute value function.
\[I = \int\limits_0^{\dfrac{1}{{\sqrt 3 }}} { - \left( {3{x^2} - 1} \right)} dx + \int\limits_{\dfrac{1}{{\sqrt 3 }}}^1 {\left( {3{x^2} - 1} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^{\dfrac{1}{{\sqrt 3 }}} {\left( {1 - 3{x^2}} \right)} dx + \int\limits_{\dfrac{1}{{\sqrt 3 }}}^1 {\left( {3{x^2} - 1} \right)} dx\]
Now solve the integrals by using the formulas \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\] and \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\].
\[ \Rightarrow I = \left[ {x - \dfrac{{3{x^3}}}{3}} \right]_0^{\dfrac{1}{{\sqrt 3 }}} + \left[ {\dfrac{{3{x^3}}}{3} - x} \right]_{\dfrac{1}{{\sqrt 3 }}}^1\]
\[ \Rightarrow I = \left[ {x - {x^3}} \right]_0^{\dfrac{1}{{\sqrt 3 }}} + \left[ {{x^3} - x} \right]_{\dfrac{1}{{\sqrt 3 }}}^1\]
Apply the upper and lower limits.
\[ \Rightarrow I = \left[ {\left( {\dfrac{1}{{\sqrt 3 }} - {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^3}} \right) - \left( {0 - {0^3}} \right)} \right] + \left[ {\left( {{1^3} - 1} \right) - \left( {{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^3} - \dfrac{1}{{\sqrt 3 }}} \right)} \right]\]
\[ \Rightarrow I = \dfrac{1}{{\sqrt 3 }} - \dfrac{1}{{3\sqrt 3 }} + \left[ {\left( {1 - 1} \right) - \left( {\dfrac{1}{{3\sqrt 3 }} - \dfrac{1}{{\sqrt 3 }}} \right)} \right]\]
\[ \Rightarrow I = \dfrac{1}{{\sqrt 3 }} - \dfrac{1}{{3\sqrt 3 }} - \dfrac{1}{{3\sqrt 3 }} + \dfrac{1}{{\sqrt 3 }}\]
\[ \Rightarrow I = \dfrac{2}{{\sqrt 3 }} - \dfrac{2}{{3\sqrt 3 }}\]
\[ \Rightarrow I = \dfrac{2}{{\sqrt 3 }}\left( {1 - \dfrac{1}{3}} \right)\]
\[ \Rightarrow I = \dfrac{2}{{\sqrt 3 }}\left( {\dfrac{2}{3}} \right)\]
\[ \Rightarrow I = \dfrac{4}{{3\sqrt 3 }}\]
Thus, \[\int\limits_0^1 {\left| {3{x^2} - 1} \right|} dx = \dfrac{4}{{3\sqrt 3 }}\].



Option ‘B’ is correct



Note: Students often make mistakes while calculating the intervals of the absolute value function. So, to calculate the interval substitute the given function in the absolute value function formula and simplify the inequality equations.