Which integral gives the area of the region in the first quadrant bounded by the axes, y = e^x, x = e^y, and the line x = 4?
The answer is an integral.
I know y=e^x has no area bounded, but I don't know how to incorporate it all.
What's the problem? A straightforward integration. The only wrinkle arises because the boundary changes at x=1. The area is
∫[0,1] e^x dx + ∫[1,4] e^x - ln(x) dx
OH I completely forgot to do the "+". I kept typing in the answer without the "+" and kept getting it wrong and decided to get help on the last try. Thank you very much Steve.
good. I guess we could also have written it as
∫[0,4] e^x - ∫[1,4] ln(x) dx
To find the area of the region in the first quadrant bounded by the axes, y = e^x, x = e^y, and the line x = 4, you can break down the region into two parts:
1. The region between the y-axis and the curve y = e^x.
2. The region between the line x = e^y and the line x = 4.
Let's calculate the area of each of these regions and add them together.
1. The region between the y-axis and the curve y = e^x:
To find the area of this region, integrate the function y = e^x with respect to x, from x = 0 to x = ln(4). This integrates the function over the values of x where it is between the y-axis and the curve.
∫(0 to ln(4)) e^x dx
2. The region between the line x = e^y and the line x = 4:
To find the area of this region, integrate the constant function x = 4 with respect to y, from y = 0 to y = ln(4). This integrates the constant value of x between the two curves.
∫(0 to ln(4)) 4 dy
Finally, add the areas of these two regions together to find the total area:
∫(0 to ln(4)) e^x dx + ∫(0 to ln(4)) 4 dy
This is the integral that gives the area of the region in the first quadrant bounded by the axes, y = e^x, x = e^y, and the line x = 4.