(a) Compute the area of the bounded region enclosed by the curve y = e^x, the line y = 12, and the y-axis.
(b) How does this area compare with the value of the integral ∫1-12(ln x dx)? Explain your answer. (A picture may be helpful.)
intersection of y = 12 and y = e^x
e^x = 12
x = ln 12
area = ?(12 - e^x)dx from 0 to ln12
= [12x - e^x] from 0 to ln12
= 12ln12 - e^(ln12) - (0 - e^0)
= 12ln12 - 12 - 0 + 1
= 12ln12 - 11
http://www.wolframalpha.com/input/?i=area+between+y+%3D+e%5Ex,++y+%3D+12+from+0+to+ln12
To compute the area of the bounded region enclosed by the curve y = e^x, the line y = 12, and the y-axis, we can break it into two parts:
Part 1: Determine the x-coordinate where the curve y = e^x intersects with the line y = 12.
To find this intersection point, we set the two equations equal to each other:
e^x = 12
Taking the natural logarithm of both sides:
ln(e^x) = ln(12)
x = ln(12)
So the curve y = e^x intersects the line y = 12 at the point (ln(12), 12).
Part 2: Compute the area between the curve y = e^x, the line y = 12, and the y-axis from x = 0 to x = ln(12).
The area can be found using the following integral:
∫[0, ln(12)] (12 - e^x) dx.
To evaluate this integral, we subtract the integral of e^x from 12 integrated from 0 to ln(12).
∫[0, ln(12)] (12 - e^x) dx = 12x - ∫[0, ln(12)] e^x dx.
To evaluate the integral of e^x, we take the antiderivative, which is simply e^x.
Therefore, the area of the bounded region is:
12ln(12) - (e^ln(12) - e^0)
Simplifying further:
12ln(12) - (12 - 1)
12ln(12) - 11
So, the area of the bounded region enclosed by the curve y = e^x, the line y=12, and the y-axis is 12ln(12) - 11.
Now, let's compare this area with the value of the integral ∫[1, 12] (ln x) dx.
This integral represents the area between the curve y = ln(x), the vertical lines x = 1 and x = 12, and the x-axis.
To evaluate this integral, we use the fundamental theorem of calculus:
∫[1, 12] (ln x) dx = [x(ln x - 1)]|[1, 12]
Plugging in the values:
(12 ln 12 - 11) - (1 ln 1 - 0)
Simplifying further:
12 ln 12 - 11
Thus, we can see that the area enclosed by the curve y = e^x, the line y = 12, and the y-axis (12ln(12) - 11) is exactly the same as the value of the integral ∫[1, 12] (ln x) dx. This means that both areas are equal.
A picture would indeed be helpful to visualize this situation: