find the area of the region enclosed by y=e^2x, y=e^5x, and x=1.
i am having trouble seeing how this region looks like and having problems with how to start the problem in general.
Much help is appreciated.
Below assumes:
y = e^2x means e^(2x)
and
y = e^5x means e^(5x)
The two curves (equations) intersect at x=0, y=1. Then, going in the positive x direction, they diverge. But, the right side of the region is bounded by a line at x=1.
So, find the area from x = 0 to x = 1 under the curve y = e^(5x) then subtract the area under the curve y = e^(2x) from x = 0 to x = 1.
Just to be clear, it is a vertical line at x=1.
To find the area of the region enclosed by the curves y=e^2x, y=e^5x, and x=1, we can start by visualizing the region.
First, let's look at the equations y=e^2x and y=e^5x individually.
The curve y=e^2x represents an exponential function with base e and a rate of change of 2. It starts at the point (0,1) and increases rapidly as x increases.
Similarly, the curve y=e^5x represents an exponential function with base e and a rate of change of 5. It also starts at the point (0,1) but increases even more rapidly than the previous curve as x increases.
Now, let's consider the equation x=1. This is a vertical line passing through the x-coordinate 1.
To determine the region enclosed by these curves, we need to find the points of intersection between them.
To find the intersection between y=e^2x and y=e^5x, we can set the two equations equal to each other:
e^2x = e^5x
Dividing both sides by e^2x:
1 = e^3x
Taking the natural logarithm (ln) of both sides:
ln(1) = 3x ln(e)
Since ln(1) = 0, we have:
0 = 3x
So the intersection point is x = 0.
To find the intersection between y=e^2x and x=1, substitute x=1 into y=e^2x:
y = e^2(1) = e^2 = 7.39 (approx.)
Therefore, the intersection point between y=e^2x and x=1 is (1, 7.39).
To find the intersection between y=e^5x and x=1, substitute x=1 into y=e^5x:
y = e^5(1) = e^5 = 148.41 (approx.)
So, the intersection point between y=e^5x and x=1 is (1, 148.41).
Now that we have identified the points of intersection, let's plot these curves on a graph.
The curve y=e^2x starts at (0,1) and increases rapidly.
The curve y=e^5x starts at (0,1) and increases even more rapidly.
The vertical line x=1 passes through the x-coordinate 1.
Based on the values we calculated, the curve y=e^2x intersects x=1 at (1, 7.39), and the curve y=e^5x intersects x=1 at (1, 148.41).
Now, to find the area of the enclosed region, we need to integrate the difference in the y-values of the two curves with respect to x, from x=0 to x=1.
The expression for the area can be written as:
Area = ∫[0,1] (e^5x - e^2x) dx
Evaluating this integral will give us the area of the region enclosed by the curves y=e^2x, y=e^5x, and x=1.