Question

What is the value of the definite integral \[\int\limits_1^5 {\left( {\left| {x - 3} \right| + \left| {1 - x} \right|} \right)} dx\]?
A. \[10\]
B. \[\dfrac{5}{6}\]
C. \[21\]
D. \[12\]

Answer 1

Hint: Here, a definite integral with absolute function is given. First, simplify the integral by applying the sum rule of the integration. Then, simplify the integrals by using the conditions for the absolute functions. After that, solve the integrals by applying the formulas of the integration. In the end, apply the limits to calculate the required answer.



Formula Used:The sum rule of the integration: \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\]
\[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]



Complete step by step solution:The given definite integral is \[\int\limits_1^5 {\left( {\left| {x - 3} \right| + \left| {1 - x} \right|} \right)} dx\].
Let consider,
\[I = \int\limits_1^5 {\left( {\left| {x - 3} \right| + \left| {1 - x} \right|} \right)} dx\]
Apply the sum rule of the definite integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\].
\[I = \int\limits_1^5 {\left| {x - 3} \right|} dx + \int\limits_1^5 {\left| {1 - x} \right|} dx\]
Simplify the integrals by checking the values of the absolute value functions on the basis of the limits.
\[I = \int\limits_1^3 {\left| {x - 3} \right|} dx + \int\limits_3^5 {\left| {x - 3} \right|} dx + \int\limits_1^5 {\left| {1 - x} \right|} dx\]
\[ \Rightarrow I = \int\limits_1^3 { - \left( {x - 3} \right)} dx + \int\limits_3^5 {\left( {x - 3} \right)} dx + \int\limits_1^5 { - \left( {1 - x} \right)} dx\]
\[ \Rightarrow I = \int\limits_1^3 {\left( {3 - x} \right)} dx + \int\limits_3^5 {\left( {x - 3} \right)} dx + \int\limits_1^5 {\left( {x - 1} \right)} dx\]
Now apply the integration formulas \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b\] and \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
\[I = \left[ {3x - \dfrac{{{x^2}}}{2}} \right]_1^3 + \left[ {\dfrac{{{x^2}}}{2} - 3x} \right]_3^5 + \left[ {\dfrac{{{x^2}}}{2} - x} \right]_1^5\]
Apply the upper and lower limits.
\[I = \left[ {\left( {3\left( 3 \right) - \dfrac{{{3^2}}}{2}} \right) - \left( {3\left( 1 \right) - \dfrac{{{1^2}}}{2}} \right)} \right] + \left[ {\left( {\dfrac{{{5^2}}}{2} - 3\left( 5 \right)} \right) - \left( {\dfrac{{{3^2}}}{2} - 3\left( 3 \right)} \right)} \right] + \left[ {\left( {\dfrac{{{5^2}}}{2} - 5} \right) - \left( {\dfrac{{{1^2}}}{2} - 1} \right)} \right]\]
\[ \Rightarrow I = \left[ {\left( {9 - \dfrac{9}{2}} \right) - \left( {3 - \dfrac{1}{2}} \right)} \right] + \left[ {\left( {\dfrac{{25}}{2} - 15} \right) - \left( {\dfrac{9}{2} - 9} \right)} \right] + \left[ {\left( {\dfrac{{25}}{2} - 5} \right) - \left( {\dfrac{1}{2} - 1} \right)} \right]\]
\[ \Rightarrow I = \left[ {\dfrac{{18 - 9}}{2} - \dfrac{{6 - 1}}{2}} \right] + \left[ {\dfrac{{25 - 30}}{2} - \dfrac{{9 - 18}}{2}} \right] + \left[ {\dfrac{{25 - 10}}{2} - \dfrac{{1 - 2}}{2}} \right]\]
\[ \Rightarrow I = \left[ {\dfrac{9}{2} - \dfrac{5}{2}} \right] + \left[ {\dfrac{{ - 5}}{2} - \dfrac{{ - 9}}{2}} \right] + \left[ {\dfrac{{15}}{2} - \dfrac{{ - 1}}{2}} \right]\]
\[ \Rightarrow I = \left[ {\dfrac{9}{2} - \dfrac{5}{2}} \right] + \left[ {\dfrac{{ - 5}}{2} + \dfrac{9}{2}} \right] + \left[ {\dfrac{{15}}{2} + \dfrac{1}{2}} \right]\]
\[ \Rightarrow I = \dfrac{{9 - 5}}{2} + \dfrac{{ - 5 + 9}}{2} + \dfrac{{15 + 1}}{2}\]
\[ \Rightarrow I = \dfrac{4}{2} + \dfrac{4}{2} + \dfrac{{16}}{2}\]
\[ \Rightarrow I = 2 + 2 + 8\]
\[ \Rightarrow I = 12\]
Therefore, \[\int\limits_1^5 {\left( {\left| {x - 3} \right| + \left| {1 - x} \right|} \right)} dx = 12\].



Option ‘D’ is correct



Note: Students directly solve the absolute value function as the normal function. They did not check whether the function is changing its sign at some point. Because of that, they get the wrong answer. So, always check the changing point of the absolute value function.