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Evaluate:\(\int\limits_{-2}^2\frac{1}{1+e^x}dx\) |
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Answer» Let I = \(\int\limits_{-2}^2\frac{1}{1+e^x}dx\)----(1) \(\therefore\) I = \(\int\limits_{-2}^2\frac{1}{1+e^{-x}}dx\) ⇒ I = \(\int\limits_{-2}^2\frac{e^x}{1+e^x}dx\)----(2) \(\therefore\) 2I = \(\int\limits_{-2}^2\frac{1+e^x}{1+e^x}dx\) (By adding (1) and (2)) ⇒ I = \(\frac12\int\limits_{-2}^{2}dx = \frac12x^2_{-2}\) = \(\frac12(2-(-2))=\frac12\times4=2\) |
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