## Calculus 8th Edition

(a) $f(x)=e^{x}\geq 1+x$ if $x\geq 0$ (b) $\frac{4}{3}\leq \int_{0}^{1} e^{x^{2}} dx\leq e$
(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, $\frac{4}{3}\leq \int_{0}^{1} e^{x^{2}} dx\leq e$