Americas
Europe
Q20E
Expert-verifiedEvaluate the integrals.
\(\int_{ - 10}^{10} {\frac{{2{e^x}}}{{\sinh x + \cosh x}}} dx\)
The value of \(\int_{ - 10}^{10} {\frac{{2{e^x}}}{{\sinh x + \cosh x}}} dx\)is \(40\).
Integral calculus is an area of mathematics that deals with the calculation, properties, and applications of integrals.
The given data is,
\(\int_{ - 10}^{10} {\frac{{2{e^x}}}{{\sinh x + \cosh x}}} dx\)
Given that: \(\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}\) and \(\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\)
Therefore
\(\begin{aligned}{c}\sinh x + \cosh x = \frac{{{e^x} - {e^{ - x}}}}{2} + \frac{{{e^x} + {e^{ - x}}}}{2}\\ = \frac{{2{e^x}}}{2}\\ = {e^x}\\ = \int_{ - 10}^{10} {\frac{{2{e^x}}}{{{e^x}}}} dx\\ = \int_{ - 10}^{10} 2 dx\\ = \int_{ - 10}^{10} 2 \cdot {x^0}dx\end{aligned}\)
Then it is expressed as,
\(\begin{aligned}{c}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C\\ = \left( {2 \times \frac{{{x^{0 + 1}}}}{{0 + 1}}} \right)_{ - 10}^{10}\\ = (2x)_{ - 10}^{10}\\ = (2 \times 10 - 2 \times ( - 10))\\ = 20 + 20\\ = 40\end{aligned}\)
Hence the value of \(\int_{ - 10}^{10} {\frac{{2{e^x}}}{{\sinh x + \cosh x}}} dx\) is 40.
94% of StudySmarter users get better grades.
Sign up for free