The value of the definite integral \(I=\int_{-1}^{1}\displaystyle\frac{1}{1+\sqrt{e^x}}\,dx\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
Since \(\sqrt{e^x} = e^{x/2}\),
\( \displaystyle I = \int_{-1}^{1} \frac{1}{1 + e^{x/2}}\,dx \)
By the property \(\int_{-a}^{a} f(x)\,dx = \int_{-a}^{a} f(-x)\,dx\), replacing x by \(-x\):
\( \displaystyle I = \int_{-1}^{1} \frac{1}{1 + e^{-x/2}}\,dx \)
\( \displaystyle = \int_{-1}^{1} \frac{e^{x/2}}{e^{x/2} + 1}\,dx \)
Adding the two forms of I:
\( \displaystyle 2I = \int_{-1}^{1} \frac{1 + e^{x/2}}{1 + e^{x/2}}\,dx \)
\( \displaystyle = \int_{-1}^{1} 1\,dx = 2 \)
\(I = 1\)
Tip: A result worth remembering: \(\int_{-a}^{a} \frac{1}{1 + e^{kx}}\,dx = a\) for every constant k, because \(f(x) + f(-x) = 1\).