Answer
$\approx 1.9972$
Work Step by Step
We are given $f(x)=e^{\sqrt x}=e^{x^{\frac{1}{2}}}$
with $a_0=1$
The Taylor series for $e^x$
is $1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+...+{\frac{1}{n!}}x^n+...$
which converges for $x$ in $(-\infty,\infty)$; replace each $x$ with $x^{1/2}$ to get
$e^{x^{\frac{1}{2}}}=1+x^{1/2}+\frac{1}{2!}(x^{1/2})^2+\frac{1}{3!}(x^{1/2})^3+\frac{1}{4!}(x^{1/2})^4$
$=1+x^{\frac{1}{2}}+\frac{x}{2}+\frac{x^{\frac{3}{2}}}{6}+\frac{x^2}{24}$
$\int^{1}_{0}(e^{x^{\frac{1}{2}}})dx \approx \int^1_0(1+x^{\frac{1}{2}}+\frac{x}{2}+\frac{x^{\frac{3}{2}}}{6}+\frac{x^2}{24})dx$
$=(x+\frac{2x^\frac{3}{2}}{3}+\frac{x^2}{4}+\frac{x^\frac{5}{2}}{15}+\frac{x^{3}}{72})|^1_0$
$\approx 1.9972$
