Question

What is the value of the integral \[\int\limits_{ - 1}^1 {\dfrac{{\sin x - {x^2}}}{{3 - \left| x \right|}}} dx\]?
A. 0
B. \[2\int\limits_0^1 {\dfrac{{\sin x}}{{3 - \left| x \right|}}} dx\]
C. \[2\int\limits_0^1 {\dfrac{{ - {x^2}}}{{3 - \left| x \right|}}} dx\]
D. \[2\int\limits_0^1 {\dfrac{{\sin x - {x^2}}}{{3 - \left| x \right|}}} dx\]

Answer 1

Hint: Here, a definite integral is given. First, simplify the given integral by separating the terms. Then, check whether the separated functions are even or odd. If the function is odd, then the value of that integral is 0. If the function is even, then solve the integral by applying the integration rule \[\int\limits_{ - a}^a {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\]. In the end, simplify the integral to get the required answer.



Formula Used:\[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function. Means, \[f\left( { - x} \right) = - f\left( x \right)\]
\[\int\limits_{ - a}^a {f\left( x \right) dx} = 2\int\limits_0^a {f\left( x \right) dx} \], if the function \[f\left( x \right)\] is an even function. Means, \[f\left( { - x} \right) = f\left( x \right)\]
\[\sin \left( { - x} \right) = - \sin x\]



Complete step by step solution:The given integral is \[\int\limits_{ - 1}^1 {\dfrac{{\sin x - {x^2}}}{{3 - \left| x \right|}}} dx\].
Let consider,
\[I = \int\limits_{ - 1}^1 {\dfrac{{\sin x - {x^2}}}{{3 - \left| x \right|}}} dx\]
Split the integral function into two terms.
\[I = \int\limits_{ - 1}^1 {\dfrac{{\sin x}}{{3 - \left| x \right|}}} dx - \int\limits_{ - 1}^1 {\dfrac{{{x^2}}}{{3 - \left| x \right|}}} dx\]
Now let’s check whether the functions are odd or even.
Let consider, \[f\left( x \right) = \dfrac{{\sin x}}{{3 - \left| x \right|}}\]
Then, \[f\left( { - x} \right) = \dfrac{{\sin \left( { - x} \right)}}{{3 - \left| { - x} \right|}}\]
\[ \Rightarrow f\left( { - x} \right) = \dfrac{{ - \sin x}}{{3 - \left| x \right|}}\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
Therefore, \[f\left( x \right) = \dfrac{{\sin x}}{{3 - \left| x \right|}}\] is an odd function.

Also, let consider \[g\left( x \right) = \dfrac{{{x^2}}}{{3 - \left| x \right|}}\]
Then, \[g\left( { - x} \right) = \dfrac{{{{\left( { - x} \right)}^2}}}{{3 - \left| { - x} \right|}}\]
\[g\left( { - x} \right) = \dfrac{{{x^2}}}{{3 - \left| x \right|}}\]
\[ \Rightarrow g\left( { - x} \right) = g\left( x \right)\]
Therefore, \[g\left( x \right) = \dfrac{{{x^2}}}{{3 - \left| x \right|}}\] is an even function.
Now apply the properties of the definite integral \[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function and \[\int\limits_{ - a}^a {f\left( x \right) dx} = 2\int\limits_0^a {f\left( x \right) dx} \], if the function \[f\left( x \right)\] is an even function.
We get,
\[I = 0 - 2\int\limits_0^1 {\dfrac{{{x^2}}}{{3 - \left| x \right|}}} dx\]
\[ \Rightarrow I = - 2\int\limits_0^1 {\dfrac{{{x^2}}}{{3 - \left| x \right|}}} dx\]
\[ \Rightarrow I = 2\int\limits_0^1 {\dfrac{{ - {x^2}}}{{3 - \left| x \right|}}} dx\]
Therefore, \[\int\limits_{ - 1}^1 {\dfrac{{\sin x - {x^2}}}{{3 - \left| x \right|}}} dx = 2\int\limits_0^1 {\dfrac{{ - {x^2}}}{{3 - \left| x \right|}}} dx\].



Option ‘C’ is correct


Note: Sometimes students get confused and write \[\sin \left( { - x} \right) = \sin x\]. Which is a wrong formula. Because of that, \[\dfrac{{\sin x}}{{3 - \left| x \right|}}\] will be considered as an even function and they get a different solution. The correct formula is \[\sin \left( { - x} \right) = - \sin x\].