Question

Show that \(\int_0^\infty {{e^{ - {x^2}}}} dx = \int_0^1 {\sqrt { - \ln y} } dy\)by interrupting the integrals as areas.

Short answer

Proved

Step-by-step solution

Find the values of \(x\) and \(y\)

The equation of the curve is shown as follows:

\(y = {e^{ - {x^2}}} \ldots \ldots (1)\)

The area under the curve \(y = {e^{ - {x^2}}}\)is shown as follows:

\({A_1} = \int_0^\infty {{e^{ - {x^2}}}} dx \ldots \ldots (2)\)

Here, the values of \(x\)ranges between \(0 \le x < \infty \) and \(y\) ranges between \(0 < y \le 1\).

Calculate area using equations within limits

Show the equation of the curve.

\(y = {e^{ - {x^2}}}\)

Modify equation (1).

Take ln on both sides of equation (1).

\(\begin{aligned}{c}\ln \;y &= \ln \left( {{e^{ - {x^2}}}} \right)\\\ln \;y &= - {x^2}\\ - \ln \;y &= {x^2}\\x &= \sqrt { - \ln \;y} \ldots \ldots \left( 3 \right)\end{aligned}\)

Calculate the area using equation (3), within limits\((0,1)\).

Consider the area obtained using equation (4) within limits \((0,1)\)is \({\rm{A2}}\).

\({A_2} = \int_0^1 {\sqrt { - \ln y} } dy \ldots \ldots {\rm{ }}\left( 4 \right)\)

Compare equations (2) and (4).

Obtain \({\rm{A1}}\)and \({\rm{A2}}\)represents the same area.

\(\begin{aligned}{c}{\rm{A1 = A2}}\\\int_0^\infty {{e^{ - {x^2}}}} dx = \int_0^1 {\sqrt { - \ln \;y} } dy\end{aligned}\)

Therefore, the integral function\(\int_0^\infty {{e^{ - {x^2}}}} dx = \int_0^1 {\sqrt { - \ln \;y} } dy\)is proved.