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Find the Maclaurin series for the functions $e^{x}$ and $\sin x$, and hence expand $e^{\sin x}$ up to the term in $x^{4} .$ Hence integrate $\int_{0}^{1} e^{\sin x} d x$.
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$e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots$
$\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots$
so that

$e^{\sin x}=1+\sin x+\frac{\sin ^{2} x}{2 !}+\frac{\sin ^{3} x}{3 !}+\frac{\sin ^{4} x}{4 !}+\cdots$
$=1+\left(x-\frac{x^{3}}{3 !}+\cdots\right)+\frac{\left(x-\frac{x^{3}}{3 !}+\cdots\right)^{2}}{2 !}+\frac{\left(x-\frac{x^{3}}{3 !}+\cdots\right)^{3}}{3 !}+\frac{\left(x-\frac{x^{3}}{3 !}+\cdots\right)^{4}}{4 !}+\cdots$
$=1+x-\frac{x^{3}}{3 !}+\cdots+\frac{x^{2}-\frac{2 x^{4}}{3 !}+\cdots}{2 !}+\frac{x^{3}+\cdots}{3 !}+\frac{x^{4}+\cdots}{4 !}+\cdots$
$=1+x-\frac{x^{3}}{3 !}+\frac{x^{2}}{2 !}-\frac{2 x^{4}}{2 ! 3 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\cdots$
$=1+x+\frac{x^{2}}{2}-\frac{x^{4}}{8}+\cdots$
Hence $\int_{0}^{1} e^{\sin x} d x=\int_{0}^{1}\left(1+x+\frac{x^{2}}{2}-\frac{x^{4}}{8}+\cdots\right) d x=x+\frac{x^{2}}{2}+\frac{x^{3}}{6}-\left.\frac{x^{5}}{40}\right|_{0} ^{1}=\frac{394}{240} \simeq 1.6417 . \quad$ The $\quad$ GDC
gives $\int_{0}^{1} e^{\sin x} d x=1.63187$
by Diamond (79,336 points)

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