Question

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

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 them to get the required answer.



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



Complete step by step solution:The given definite integral is \[\int\limits_{ - 1}^2 {\left| x \right|} dx\].
Let consider,
\[I = \int\limits_{ - 1}^2 {\left| x \right|} dx\]
Break the integration by using absolute function \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\].
We get,
\[I = \int\limits_{ - 1}^0 {\left( { - x} \right)} dx + \int\limits_0^2 x dx\]
\[ \Rightarrow I = - \int\limits_{ - 1}^0 x dx + \int\limits_0^2 x dx\]
Solve the integrals by applying the integration formula \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
\[ \Rightarrow I = - \left[ {\dfrac{{{x^2}}}{2}} \right]_{ - 1}^0 + \left[ {\dfrac{{{x^2}}}{2}} \right]_0^2\]
Apply the upper and lower limit.
\[ \Rightarrow I = - \left[ {\dfrac{{{0^2}}}{2} - \dfrac{{{{\left( { - 1} \right)}^2}}}{2}} \right] + \left[ {\dfrac{{{2^2}}}{2} - \dfrac{{{0^2}}}{2}} \right]\]
\[ \Rightarrow I = - \left[ { - \dfrac{1}{2}} \right] + \dfrac{4}{2}\]
\[ \Rightarrow I = \dfrac{1}{2} + \dfrac{4}{2}\]
\[ \Rightarrow I = \dfrac{5}{2}\]
Thus, \[\int\limits_{ - 1}^2 {\left| x \right|} dx = \dfrac{5}{2}\].



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