A region is bounded in the second quadrant by the curve y = ln(1–x), the line y=3, and the y-axis. Find the area of the region.
If you draw this you see you want the integral from y =0 to y = 3 of -x dy
y = ln(1-x)
e^y = 1-x
-x = e^y -1
so integral from 0 to 3
e^y dy - dy
e^y - y at 3 = e^3-3
e^y - y at 0 = 1
so
e^3 - 4
about 20 - 4 = 16
Wolfram amazes me more every day.
http://www.wolframalpha.com/input/?i=area+between+y%3D+ln%281-x%29+%2C+y%3D3+%2C+x%3D0
notice that Damon took horiontal slices, while Wolfram took vertical slices.
Same result of course.
To find the area of the region bounded in the second quadrant by the curve y = ln(1–x), the line y = 3, and the y-axis, we need to calculate the area between the curve and the line.
Here's how you can solve it step by step:
1. Identify the points of intersection between the curve and the line. In this case, we need to find where y = ln(1–x) intersects with y = 3.
Set the equations equal to each other:
ln(1–x) = 3
2. Solve for x by taking the exponential of both sides of the equation:
e^(ln(1–x)) = e^3
This simplifies to:
1 – x = e^3
3. Solve for x by isolating it on one side of the equation:
x = 1 – e^3
4. Now we need to determine the limits of integration. Since the region is bounded by the y-axis, we integrate from x = 0 to x = 1 – e^3, since that's the point of intersection.
5. Set up the integral for calculating the area:
area = ∫[0 to (1 – e^3)] (upper curve - lower curve) dx,
where upper curve = ln(1–x) and lower curve = 3
6. Integrate the expression:
area = ∫[0 to (1 – e^3)] [ln(1–x) - 3] dx
7. Evaluate the integral. Taking the integral of ln(1–x) and 3 gives us:
area = [xln(1–x) - 3x] evaluated from 0 to (1–e^3)
8. Substitute the limits of integration and solve:
area = [(1 – e^3)ln(1–(1–e^3)) - 3(1–e^3)] - [0 - 0]
Simplifying further will give you the exact numerical value for the area of the region bounded by the given curve and line in the second quadrant.