Question

The value of the integral ${\int }_0^1{e}^{x^2}dx$

• A. Is less than $1$
• B. Is greater than $1$
• C. Is less than or equal to $1$
• D. Lies in the closed interval $[1,e]$

Answer 1

The correct option is D Lies in the closed interval $[1,e]$

${\int }_0^1{e}^{x^2}dx$ $={\int }_0^1\sum _{n=0}^{\infty }\frac{x^{2n}}{n!}dx$

$=\sum _{n=0}^{\infty }\frac{1}{n!}{\int }_0^1{x}^{2n}dx$

$=\sum _{n=0}^{\infty }\frac{1}{n!(2n+1)}$

$=1+\frac{1}{3}+\frac{1}{10}+\frac{1}{42}+....$

$=\frac{1}{216}+\frac{1}{1320}+\frac{1}{9360}+....$

$\approx 1.462$