Answer
(a) $f(x)=e^{x}\geq 1+x$ if $x\geq 0$
(b) $\frac{4}{3}\leq \int_{0}^{1} e^{x^{2}} dx\leq e$
Work Step by Step
(a) As we are given $f(x)=e^{x}\geq 1+x$ if $x\geq 0$
Prove this inequality .
Consider $f(x)=e^{x}- (1+x)$
Now, $f'(x)=e^{x}- 1$
$f'(0)=e^{0}- 1=0$
Since, $f(x)$ is increasing on $[0,\infty)$
Therefore, $f(x)\geq 0$ for $x\geq 0$
Thus, $e^{x}- (1+x)\geq 0$
This implies
$e^{x}\geq (1+x)$ if $x\geq 0$
(b) Deduce that $\frac{4}{3}\leq \int_{0}^{1} e^{x^{2}} dx\leq e$
Therefore, $f(x)\geq 0$ for $x\geq 0$
$e^{x^{2}}- (1+x^{2})\geq 0$
Thus, $e^{x^{2}}\geq (1+x^{2})$
For $0\leq x\leq 1$
$(1+x^{2})\leq e^{x^{2}}\leq e^{x}$
Further,
$\int_{0}^{1} (1+x^{2}) dx\leq \int_{0}^{1} e^{x^{2}}dx \leq \int_{0}^{1} e^{x} dx$
Hence, $\frac{4}{3}\leq \int_{0}^{1} e^{x^{2}} dx\leq e$
