Question

If \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{e^{\cos x}}\sin x,}&{\left| x \right| \le 2}\\{2,}&{{\rm{otherwise}}}\end{array}} \right.\], then find the value of \[\int_{ - 2}^3 {f\left( x \right)dx} \].
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: First we will find the simplest form of the \[\left| x \right| \le 2\]. Then we will break the integration as a sum of two integrations. The limits of the first integration will be -2 to 2 and the function will be \[{e^{\cos x}}\sin x\] as \[\left| x \right| \le 2\]. The limits of the second integration will be 2 to 3. Then we will check whether \[{e^{\cos x}}\sin x\] is an odd function or an even function. According to that, we apply the property of definite integral and also integrate the second integration.



Formula Used:Definite integral property:
\[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\]



Complete step by step solution:We know that if \[\left| x \right| \le a\] then \[ - a \le x \le a\].
We can rewrite \[\left| x \right| \le 2\] as \[ - 2 \le x \le 2\].
Now rewrite the given information that is \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{e^{\cos x}}\sin x,}&{\left| x \right| \le 2}\\{2,}&{{\rm{otherwise}}}\end{array}} \right.\].
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{e^{\cos x}}\sin x,}&{ - 2 \le x \le 2}\\{2,}&{{\rm{otherwise}}}\end{array}} \right.\]
It means when \[ - 2 \le x \le 2\], then \[f\left( x \right) = {e^{\cos x}}\sin x\].
And when \[x \notin \left[ { - 2,2} \right]\], then \[f\left( x \right) = 2\]
We will break the integration as a sum of two integrations. The limit of the first integration is -2 to 2 and the second integration is 2 to 3
Now we will the limits of the given integration,
\[\int_{ - 2}^3 {f\left( x \right)dx} \]
\[ = \int_{ - 2}^2 {f\left( x \right)dx} + \int_2^3 {f\left( x \right)dx} \]
Now substitute the value of \[f\left( x \right)\]:
\[ = \int_{ - 2}^2 {{e^{\cos x}}\sin xdx} + \int_2^3 {2dx} \] ……(i)
Assume that \[g\left( x \right) = {e^{\cos x}}\sin x\]
Now putting x = -x
\[g\left( { - x} \right) = {e^{\cos \left( { - x} \right)}}\sin \left( { - x} \right)\]
\[ \Rightarrow g\left( { - x} \right) = - {e^{\cos x}}\sin x\]
\[ \Rightarrow g\left( { - x} \right) = - g\left( x \right)\]
Therefore \[g\left( x \right)\] is an odd function.
According the definite integral property \[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\]:
\[\int_{ - 2}^2 {{e^{\cos x}}\sin xdx} = 0\]
Substitute \[\int_{ - 2}^2 {{e^{\cos x}}\sin xdx} = 0\] in equation (i):
\[\int_{ - 2}^2 {{e^{\cos x}}\sin xdx} + \int_2^3 {2dx} \]
\[ = 0 + \int_2^3 {2dx} \]
Integration the second integration:
\[ = 0 + 2\left[ x \right]_2^3\]
\[ = 0 + 2\left[ {3 - 2} \right]\]
\[ = 0 + 2 \cdot 1\]
\[ = 2\]



Option ‘C’ is correct



Note: Students try to solve the integration \[\int_{ - 2}^2 {{e^{\cos x}}\sin xdx} \] by using the substitution method. But it is a bit lengthy process. We can easily solve it by using the property of definite integral.