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$\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x=$
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Verified Answer
The correct answer is:
$e-\frac{1}{e}$
$\begin{aligned} \text{Let} \quad I &=\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x \\ & \Rightarrow \quad I=\int_{-1}^{1} x^{27} \cos x d x+\int_{-1}^{1} e^{x} d x \end{aligned}$
Since, $f(x)=x^{27} \cos x$ is an odd function.
$$
\left\{\because \int_{-a}^{a} f(x) d x=0 \text { if } f(x) \text { is odd function }\right\}
$$
So, $\quad I=0+\left[e^{x}\right]_{-1}^{1}=e-\frac{1}{e}$
Since, $f(x)=x^{27} \cos x$ is an odd function.
$$
\left\{\because \int_{-a}^{a} f(x) d x=0 \text { if } f(x) \text { is odd function }\right\}
$$
So, $\quad I=0+\left[e^{x}\right]_{-1}^{1}=e-\frac{1}{e}$
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