How Does Using Midpoints in Integrals Estimate Area?

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1. let A be the area of the region that lies
under the graph of f(x) = e^-x , between x=0 and x=2.

1-Estimate the area using the sample point to be the midpoint and using fuor sub-intervals.
x*1=(0+0.5)\2=1\4.
x*2=(1+0.5)\2=3\4.
x*3=(1+3\2)\2=5\4.
x*4=(2+3\2)\2=7\4
A=segma i=1 to n f(x*i) X delta x =0.8357

I copied the answer from the board with my doctor but I didn't understand the solution
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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