Question

$\int\limits_{-1}^1 (x^{27} \, \cos \, x + e^x) dx = $

• A. $\frac{2e - 1}{e}$
• B. $\frac{e + 1}{e}$
• C. $e - \frac{1}{e}$
• D. $\frac{1}{e}$

Answer 1

Answer: (C)

Let $ I = \int\limits_{-1}^1 (x^{27} \, \cos \, x + e^x) dx $
$\Rightarrow \:\:\: I = \int\limits_{-1}^1 x^{27} \, \cos \, x\, dx + \int\limits_{-1}^1 e^x \, dx $
Since $f(x) = x^{27} \, \cos \, x$ is an odd function.
$\therefore \:\:\: \int\limits_{-a}^a f(x) dx = 0$
So, $I = 0 + |e^x |_{-1}^{1} = e - \frac{1}{e} $