1

Evaluate the value of: $$\int_0^1 e^{\left(x^2\right)}~dx+\int_1^e\sqrt{\ln x}~dx$$

Here is what I have tried:

$\ln x >\sqrt{\ln x}$ when $x=[1,e]$

$e^x>e^{\left(x^2\right)}$ when $x=[0,1]$

And after that put under the integral and obtain:

$$\int_0^1 e^{\left(x^2\right)}~dx+\int_1^e\sqrt{\ln x}~dx\leq e$$

And it is easy to see:

$$e\geq \int_0^1 e^{\left(x^2\right)}~dx + \int_1^e\sqrt{\ln x}~dx>1$$

But it is not solved just I know is smaller than $e$ and bigger than $1$

I can write a series but it will only give me an approximation.

projectilemotion
  • 15,428
Ica Sandu
  • 565

3 Answers3

5

HINT: If $f$ is an increasing function on the interval $[a,b]$ and $f^{-1}$ is its inverse function, then

$$ \int_{a}^{b} f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = b\,f(b)-a\, f(a). \tag{1}$$

Jack D'Aurizio
  • 353,855
2

Note that we can obtain a relationship between the two integrals via the substitution $x=e^{u^2}$ and integration by parts: $$\begin{align}\int_1^e \sqrt{\ln{x}}~dx&=\int_0^1 u\cdot 2ue^{u^2}~du\\&=\left[e^{u^2}u\right]_0^1-\int_0^1 e^{u^2}~du\\&=e-\int_0^1 e^{u^2}~du \tag{1}\end{align}$$ Hence, we deduce from $(1)$ that: $$\bbox[5px,border:2px solid #C0A000]{\int_0^1 e^{x^2}~dx+\int_1^e \sqrt{\ln{x}}~dx=e}$$

projectilemotion
  • 15,428
0

$\int\limits_0^1 e^{x^2}\ dx = \frac{1}{2} \sqrt{\pi } \text{erfi}(1)$

and

$\int\limits_1^e \sqrt{\ln (x)}\ dx = e-\frac{1}{2} \sqrt{\pi } \text{erfi}(1)$ (as projectilemotion showed)...

so the sum equals $e$.

David G. Stork
  • 29,774