Inequality in Integral Show that $\dfrac{28}{81}<\int_0^\frac{1}{3}e^{x^2}dx<\dfrac{3}{8}$.
It would be great if a solution based on the Mean Value Theorem for Integrals is posted.

To obtain the upper bound, note that for $|x| < 1$
$$e^{x^2} = \sum_{k=0}^{\infty}\frac{x^{2k}}{k!} < \sum_{k=0}^{\infty}x^{2k} = \frac1{1-x^2}.$$
Hence,
$$\int_0^{1/3}e^{x^2}\,dx < \int_0^{1/3}\frac{dx}{1-x^2} <  \frac1{3}\frac{1}{1-(1/3)^2}= \frac1{3}\frac{9}{8}= \frac{3}{8}.$$
The lower bound is easily derived by truncating the Taylor series.
