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.
length of side of square = e^x
area of square = (e^x)^2 = e^2x
volume = integral dx e^2x from x = 0 to x = 2
= (1/2)e^2x at x = 2 - at x = 0
= (1/2)e^4 -(1/2)(1)
= (1/2)(e^4 -1)
To compute the volume exactly, we need to set up a definite integral to integrate over the given domain.
The volume of each individual square can be calculated by taking the width of the square, which is dx, and multiplying it by the area of the square, which is the square of the function value, f(x)^2 = (e^x)^2 = e^(2x).
Therefore, the volume of the square at a given value of x is given by dV = e^(2x) * dx.
To find the total volume, we need to integrate this expression over the domain of 0 < x < 2:
V = ∫(from 0 to 2) e^(2x) dx.
Now, to evaluate this integral, we can use the power rule of integration. The integral of e^(2x) is (1/2) * e^(2x), and applying the limits of integration, we get:
V = (1/2) * e^(2x) |(from 0 to 2).
Substituting the limits into the expression:
V = (1/2) * (e^(4) - e^(0)).
Simplifying further:
V = (1/2) * (e^(4) - 1).
Thus, the integral to compute the volume is:
V = (1/2) * (e^(4) - 1).