Question

What is the value of the definite integration \[\int_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} \]?
A. \[\dfrac{1}{3}\]
B. \[\dfrac{{14}}{3}\]
C. \[\dfrac{7}{3}\]
D. \[\dfrac{{28}}{3}\]

Answer 1

Hint: First we will find the interval where the value \[\left| {1 - {x^2}} \right|\] is positive and negative. Then rewrite the integration as sum of integration. After that integrate it to get the value of \[\int_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} \].



Formula Used:Property of definite integration:
\[\int_b^a {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} \] where \[b < c < a\]



Complete step by step solution:Given definite integration is \[\int_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} \]
Case 1: When \[1 - {x^2} < 0\]
\[1 - {x^2} < 0\]
Subtract from 1 both sides
\[ \Rightarrow - {x^2} < - 1\]
\[ \Rightarrow {x^2} > 1\]
\[ \Rightarrow x > 1\] or \[x < - 1\]
Case 1: When \[1 - {x^2} > 0\]
\[1 - {x^2} > 0\]
Subtract from 1 both sides
\[ \Rightarrow - {x^2} > - 1\]
\[ \Rightarrow {x^2} < 1\]
\[ \Rightarrow - 1 < x < 1\]
Now we will decide the positive and negative of \[\left| {1 - {x^2}} \right|\] in the interval \[\left( { - 2,3} \right)\]
We know that the value of \[1 - {x^2}\] is less than zero when \[x > 1\] or \[x < - 1\].
In other words \[1 - {x^2}\] is less than zero when \[1 < x < 3\] or \[ - 2 < x < - 1\]
We know that the value of \[1 - {x^2}\] is greater than zero when \[ - 1 < x < 1\]
Now we will rewrite the given definite integral as sum of three integrations. The limits of the first integration will be -2 to -1, limits of the second integration will be -1 to 1, and limits of the third integration will be 1 to 3.
\[\int_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} \]
\[ = \int_{ - 2}^{ - 1} { - \left( {1 - {x^2}} \right)dx} + \int_{ - 1}^1 {\left( {1 - {x^2}} \right)dx} + \int_1^3 { - \left( {1 - {x^2}} \right)dx} \]
Now applying integration formula:
\[ = - \left[ {x - \dfrac{{{x^3}}}{3}} \right]_{ - 2}^{ - 1} + \left[ {x - \dfrac{{{x^3}}}{3}} \right]_{ - 1}^1 + \left[ { - x + \dfrac{{{x^3}}}{3}} \right]_1^3\]
\[ = - \left[ { - 1 - \dfrac{{{{\left( { - 1} \right)}^3}}}{3} - \left( { - 2 - \dfrac{{{{\left( { - 2} \right)}^3}}}{3}} \right)} \right] + \left[ {1 - \dfrac{{{1^3}}}{3} - \left( { - 1 - \dfrac{{{{\left( { - 1} \right)}^3}}}{3}} \right)} \right] + \left[ { - 3 + \dfrac{{{3^3}}}{3} - \left( { - 1 + \dfrac{{{1^3}}}{3}} \right)} \right]\]
\[ = - \left[ { - 1 + \dfrac{1}{3} + 2 - \dfrac{8}{3}} \right] + \left[ {1 - \dfrac{{{1^3}}}{3} + 1 - \dfrac{1}{3}} \right] + \left[ { - 3 + 9 + 1 - \dfrac{1}{3}} \right]\]
\[ = - \left[ {1 - \dfrac{7}{3}} \right] + \left[ {2 - \dfrac{2}{3}} \right] + \left[ {7 - \dfrac{1}{3}} \right]\]
\[ = - 1 + \dfrac{7}{3} + 2 - \dfrac{2}{3} + 7 - \dfrac{1}{3}\]
\[ = 8 + \dfrac{4}{3}\]
\[ = \dfrac{{28}}{3}\]



Option ‘D’ is correct



Note: Students often make mistake to solve the given integration. They do not consider the negative value of \[\left| {1 - {x^2}} \right|\]. They integrate \[\int_{ - 2}^3 {\left( {1 - {x^2}} \right)dx} \] and get incorrect solution. Here we find the intervals where \[\left| {1 - {x^2}} \right|\] is positive and negative. After that rewrite the integration as sum of integrations and solve it to reach correct solution.