2. f(x)=e^x for 0<x<2. Imagine that you have graphed this function in the x-y plane. For each value of x in the domain, erect a square above the curve perpendicular to the x-axis and parallel to the y-z plane where one side runs from the x-axis to the curve. So the square at x=.9 has vertices (.9,0,0), (.9,e^.9,0), (.9,e^.9,e^.9), (.9,0,e^.9). All of these squares together create a volume. Set up an integral to compute the volume exactly.

To compute the volume exactly, we need to integrate the area of each square from x = 0 to x = 2.

Let's first find the area of each square. Since the squares are perpendicular to the x-axis, their side lengths are equal to the function value at each x.

For a given x, the side length of the square is e^x. The area of a square is equal to the side length squared, so the area of the square at x is (e^x)^2 = e^(2x).

To compute the volume, we need to sum up the areas of these squares from x = 0 to x = 2.

The integral to compute the volume is given by:

∫[0,2] e^(2x) dx

To solve this integral, we can use the power rule for integration.

Applying the power rule, we increase the exponent by 1 and divide by the new exponent. So integrating e^(2x) with respect to x gives us:

(1/2) * e^(2x)

To evaluate the definite integral from x = 0 to x = 2, we subtract the value of the antiderivative at the lower limit from the value at the upper limit:

[ (1/2) * e^(2 * 2) ] - [ (1/2) * e^(2 * 0) ]

Simplifying further:

[ (1/2) * e^4 ] - [ (1/2) * e^0 ]

Since e^0 is equal to 1, the expression further simplifies to:

(1/2) * e^4 - (1/2)

This gives us the exact volume of the region enclosed by the squares erected above the curve f(x) = e^x for 0 < x < 2.