SOLUTION: Sketch the region enclosed by y = e^{3x}, y=e^{7x} and x=1. Decide whether to integrate with respect to x or y, and then find the area of the region.

Algebra ->  Trigonometry-basics -> SOLUTION: Sketch the region enclosed by y = e^{3x}, y=e^{7x} and x=1. Decide whether to integrate with respect to x or y, and then find the area of the region.       Log On


   

Question 574997: Sketch the region enclosed by y = e^{3x}, y=e^{7x} and x=1.
Decide whether to integrate with respect to x or y, and then find the area of the region.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
At x=0, y=1 for both functions, so they intersect at (0,1). The region would have vertical line x=1 and the two functions as boundaries. The region has the graph of y=e%5E%287x%29 as the upper boundary for y, and the graph of y=e%5E%283x%29 as its lower boundary throughout the [0,1] interval.
Integrating with respect to x would simply mean
int%28+%28e%5E%287x%29-e%5E%283x%29%29%2C+dx%2C+0%2C+1+%29
If we try to integrate with respect to y, we need to do some calculations first:
y=e%5E%283x%29 and x=1 intersect at (1,e%5E3), and
y=e%5E%287x%29 and x=1 intersect at (1,e%5E7)
The y values for the region range between 1 and e%5E7
The inverse of y=e%5E%287x%29 is x=ln%28y%29%2F7 , which is the lower boundary for x values in the region.
On the other hand, the upper boundary for x values is x=ln%28y%29%2F3 between 1 and e%5E3), and x=1 between e%5E3) and e%5E7).
So after all those calculations, we would end up with two integrals.
int%28+%28e%5E%287x%29-e%5E%283x%29%29%2C+dx%29=e%5E%287x%29%2F7-e%5E%283x%29%2F3, so
= approx. 150.157