AFirst find the intersections of \(y=e^{2x-x^2}\) and \(y=2\):
\(2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}\)
So the area of \(R\) is given by
\(\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx\)
If you're not familiar with the error function \(\mathrm{erf}(x)\), then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.
(b) Find the intersections of the line \(y=1\) with \(y=e^{2x-x^2}\).
\(1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2\)
So the area of \(S\) is given by
\(\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx\)
\(\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx\)
which is approximately 1.546.
(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve \(y=e^{2x-x^2}\) and the line \(y=1\), or \(e^{2x-x^2}-1\). The area of any such circle is \(\pi\) times the square of its radius. Since the curve intersects the axis of revolution at \(x=0\) and \(x=2\), the volume would be given by
\(\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx\)
