Question

The value of \[\int\limits_{-1}^{1}{\dfrac{{{e}^{\left| x \right|}}}{1+{{2}^{x}}}}dx\] equals
A. e
B. \[{{e}^{2}}\]
C. e-1
D. None of these

Answer 1

Hint: The most important formula that we can use in this question is as follows
\[\int\limits_{-a}^{a}{f(x)}dx=\int\limits_{0}^{a}{\left( f(x)+f(-x) \right)}dx\]
(We are able to use the above property of definite integration in cases where the limits are symmetric about the origin or zero.)
We could use the above mentioned property of definite integration in this question because the limits in this question are symmetrical.

Complete step-by-step answer:

As mentioned in the question, we have to find the value of the given definite integral.
Now, we could use the information or the property that is mentioned in the hint to evaluate the given definite integral as follows
=\[\begin{align}
  & =\int\limits_{-1}^{1}{\dfrac{{{e}^{\left| x \right|}}}{1+{{2}^{x}}}}dx \\
 & =\int\limits_{0}^{1}{\left( \dfrac{{{e}^{\left| x \right|}}}{1+{{2}^{x}}}+\dfrac{{{e}^{\left| -x \right|}}}{1+{{2}^{-x}}} \right)}dx \\
 & =\int\limits_{0}^{1}{\left( \dfrac{{{e}^{\left| x \right|}}}{1+{{2}^{x}}}+\dfrac{{{e}^{\left| -x \right|}}}{1+\dfrac{1}{{{2}^{x}}}} \right)}dx \\
 & =\int\limits_{0}^{1}{\left( \dfrac{{{e}^{\left| x \right|}}}{1+{{2}^{x}}}+\dfrac{\left( {{2}^{x}} \right){{e}^{\left| -x \right|}}}{{{2}^{x}}+1} \right)}dx \\
 & =\int\limits_{0}^{1}{\left( \dfrac{{{e}^{x}}}{1+{{2}^{x}}}+\dfrac{\left( {{2}^{x}} \right){{e}^{-(-x)}}}{{{2}^{x}}+1} \right)}dx \\
 & =\int\limits_{0}^{1}{\left( \dfrac{{{e}^{x}}}{1+{{2}^{x}}}+\dfrac{\left( {{2}^{x}} \right){{e}^{x}}}{{{2}^{x}}+1} \right)}dx \\
 & =\int\limits_{0}^{1}{\left( \dfrac{\left( 1+{{2}^{x}} \right){{e}^{x}}}{{{2}^{x}}+1} \right)}dx \\
 & =\int\limits_{0}^{1}{{{e}^{x}}}dx \\
 & =\left[ {{e}^{x}} \right]_{0}^{1} \\
 & =e-1 \\
\end{align}\]
Hence, the value of the given definite integral is equal to e-1.

Note: The students can make an error if they don’t know the property of modulus function that is if the value of x is always greater than zero or 0, then the modulus function is removed as follows
\[\left| x \right|=x\]
And if the value of x is always lesser than zero or 0, then the modulus function is removed as follows
\[\left| x \right|=-x\]
Another most important formula that we used in this question is as follows
\[\int\limits_{-a}^{a}{f(x)}dx=\int\limits_{0}^{a}{\left( f(x)+f(-x) \right)}dx\]
(We are able to use the above property of definite integration in cases where the limits are symmetric about the origin or zero)