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Try this problem from ISI-MSQMS 2015 which involves the concept of Integral Inequality.

## INTEGRAL INEQUALITY | ISI 2015 | MSQMS | PART B | PROBLEM 7b

Show that $1<\int_{0}^{1} e^{x^{2}} d x<e$

### Key Concepts

Real Analysis

Inequality

Numbers

But Try the Problem First…

Source

ISI – MSQMS – B, 2015, Problem 7b

“INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA”

## Try with Hints

First hint

We have to show that ,

$1<\int_{0}^{1} e^{x^{2}} d x<e$

$0< x <1$

It implies, $0 < x^2 <1$

Now with this reduced form of the equation why don’t you give it a try yourself, I am sure you can do it.

Second hint

Thus, $e^0 < e^{x^2} <e^1$

i.e $1 < e^{x^2} <e$

So you are just one step away from solving your problem, go on………….

Final hint

Therefore, Integrating the inequality with limits $0$ to $1$ we get, $\int\limits_0^1 \mathrm dx < \int\limits_0^1 e^{x^2} \mathrm dx < \int\limits_0^1e \mathrm dx$