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.