Question

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

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 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 \[\int\limits_0^3 {\left| {2 - x} \right|} dx\].
Let consider,
\[I = \int\limits_0^3 {\left| {2 - x} \right|} dx\]
Substitute \[x = 2 - x\] 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,
\[\left| {2 - x} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {2 - x} \right),}&{if 2 - x < 0}\\{2 - x,}&{if 2 - x \ge 0}\end{array}} \right.\]
\[ \Rightarrow \left| {2 - x} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {2 - x} \right),}&{if x > 2}\\{2 - x,}&{if x \le 2}\end{array}} \right.\]
The interval of the integration is 0 to 3.
This implies that \[\left| {2 - x} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {2 - x} \right),}&{if 2 < x < 3}\\{2 - x,}&{if 0 < x \le 2}\end{array}} \right.\].
Break the integration by using the above absolute value function.
\[I = \int\limits_0^2 {\left( {2 - x} \right)} dx + \int\limits_2^3 { - \left( {2 - x} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^2 {\left( {2 - x} \right)} dx + \int\limits_2^3 {\left( {x - 2} \right)} dx\]
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 \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\].
\[ \Rightarrow I = \left[ {2x - \dfrac{{{x^2}}}{2}} \right]_0^2 + \left[ {\dfrac{{{x^2}}}{2} - 2x} \right]_2^3\]
Apply the upper and lower limits.
\[ \Rightarrow I = \left[ {\left( {2\left( 2 \right) - \dfrac{{{2^2}}}{2}} \right) - \left( {2\left( 0 \right) - \dfrac{{{0^2}}}{2}} \right)} \right] + \left[ {\left( {\dfrac{{{3^2}}}{2} - 2\left( 3 \right)} \right) - \left( {\dfrac{{{2^2}}}{2} - 2\left( 2 \right)} \right)} \right]\]
\[ \Rightarrow I = \left( {4 - 2} \right) + \left( {\dfrac{9}{2} - 6} \right) - \left( {2 - 4} \right)\]
\[ \Rightarrow I = 2 - \dfrac{3}{2} + 2\]
\[ \Rightarrow I = 4 - \dfrac{3}{2}\]
\[ \Rightarrow I = \dfrac{5}{2}\]
Thus, \[\int\limits_0^3 {\left| {2 - x} \right|} dx = \dfrac{5}{2}\].



Option ‘B’ is correct


Note: Students often make mistake while calculating the intervals of the absolute value function. Because of the negative sign in \[ - \left( {2 - x} \right)\], the considered this value is true for \[x \le 2\]. Which is wrong.So, calculate the interval by following the above steps.