Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

y=e^1x, y=e^4x, x=1

wrt x:

∫[0,1] e^(4x)-e^x dx
= 1/4 e^(4x) - e^x [0,1]
= (1/4 e^4 - e) - (1/4 - 1)
= 1/4 (3 + e^4 - 4e)

wrt y:

∫[1,e] lny - 1/4 lny dy + ∫[e,e^4] 1- 1/4 lny dy
= (3/4) + (1/4 (e$4 - 4e))
= 1/4 (3 + e^4 - 4e)

To sketch the region enclosed by the given curves, we need to plot the curves and identify the boundaries of the region.

The given curves are y=e^x, y=e^4x, and x=1.

Plotting these curves on a graph, we see that y=e^x is an increasing exponential curve that starts at the y-intercept (0,1), and y=e^4x is also an increasing exponential curve, but steeper than y=e^x. The vertical line x=1 passes through both curves.

To find the boundaries of the region, we need to determine the intersection points of the curves. Setting the two equations equal to each other, we have:
e^x = e^4x

Taking the natural logarithm (ln) of both sides to get rid of the exponential, we get:
x = 4x

Simplifying the equation, we have:
3x = 0
x = 0

Therefore, the two curves intersect at x = 0.

Since x=1 is a vertical line and it intersects both curves, it defines the boundaries of the region.

Now, let's decide whether to integrate with respect to x or y to find the area of the region. To do this, we need to determine which variable (x or y) is being varied as we move across the region from one boundary curve to the other.

Looking at the graph, we see that as we move from the left boundary curve (y=e^x) to the right boundary curve (y=e^4x), the y-values are changing. Therefore, we integrate with respect to y.

To find the area of the region, we integrate the difference between the upper and lower curves (y=e^4x - y=e^x) with respect to y, within the y-value range determined by the boundaries:

∫[e^x, e^4x] (e^4x - e^x) dy

We integrate from e^x to e^4x because the values of y between these two curves form the region.

To find ∫(e^4x - e^x) dy, we need to express the equation in terms of x. So we solve e^x = y for x to get x = ln(y).

Rewriting the integral with x as the variable, we have:

∫[1, e] (e^4x - e^x) dx

Evaluating this integral will give us the area of the region enclosed by the given curves.