1. Let S be the region in the first quadrant bounded by the graphs of y = e^(-x^2) y = 2x^2, and the y axis.

y x = 2 , and the y-axis.

a) Find the area of the region S.

So when I integrate, is it from 0 to .5325? I found where they intersect on the calculator. Is this the right method?

Wont there be more than one region? Have you sketched this?

Not sure what you mean. It says it's in the first quadrant.

To find the area of the region S bounded by the given curves, you can integrate the difference between the upper curve and the lower curve with respect to x over the appropriate interval.

In this case, you correctly identified the upper curve as y = 2x^2 and the lower curve as y = e^(-x^2). To find the x-values where these curves intersect, you can set them equal to each other:

2x^2 = e^(-x^2)

To solve this equation, you can use numerical methods or algebraically manipulate the equation. As you mentioned, using a calculator to find the intersection points is one method. From your calculation, it seems like the x-value of the intersection point is approximately 0.5325.

Once you have the x-value of the intersection point, you can set up the integral to find the area:

∫[a, b] (upper curve - lower curve) dx, where [a, b] is the interval over which the curves intersect.

Since you have determined that the intersection point is 0.5325, you can integrate from 0 to 0.5325:

Area = ∫[0, 0.5325] (2x^2 - e^(-x^2)) dx

Now you can evaluate this integral to find the area of region S using numerical methods or appropriate techniques like substitution or integration by parts.

Please note that the value of the integral might not be exactly representable in terms of elementary functions and may require numerical approximation methods such as using software or calculators.