Answer
(a) $e^{x}\geq (1+x+\frac{1}{2}x^{2})$ for $x\geq 0$
(b) $\int_{0}^{1} e^{x^{2}} dx=\frac{43}{30}$
Work Step by Step
(a) As we are given $f(x)=e^{x}\geq 1+x+\frac{1}{2}x^{2}$ if $x\geq 0$
Prove this inequality .
Consider $f(x)=e^{x}-[1+x+\frac{1}{2}x^{2}]$
Now, $f'(x)=e^{x}- (1+x)$
$f'(0)=e^{0}- 1=0$
Since, $f(x)$ is increasing on $[0,\infty)$
Thus, $e^{x}\geq (1+x+\frac{1}{2}x^{2})$ if $x\geq 0$
(b) Deduce that $\int_{0}^{1} e^{x^{2}} dx$
Using part (a) which is $e^{x^{2}}\geq (1+x^{2}+\frac{1}{2}x^{4})$ for $x\geq 0$, we have
$\int_{0}^{1} e^{x^{2}} dx=\int_{0}^{1}(1+x^{2}+\frac{1}{2}x^{4}) dx$
$= \int_{0}^{1}(x+\frac{x^{3}}{3}+\frac{x^{5}}{10}) dx$
$=1+\frac{1}{3}+\frac{1}{10}$
Hence, $\int_{0}^{1} e^{x^{2}} dx=\frac{43}{30}$
