Approximating the integral from 0 to 6 of (e^x dx) by 3 circumscribed rectangles of equal width on the x-axis yields ____.
a) 2e^2 + 4e^4 + 6e^6
b) 2(e^2 + e^4 + e^6)
c) 2(e + e^3 + e^5)
d) e + 3e^3 + 5e^5
e) e^2 + 3e^4 + 5e^6
sketch the graph of e^x with the rectangles of height e^2, e^4 and e^6 and base of 2 each
2 e^2 + 2 e^4 + 2 e^6
= 2 (e^2 + e^4 + e^6)
Not sure what the "circumscribed" condition means, since any rectangle can be circumscribed.
Damon's rectangles contain the approximated sections of curve within them, whatever that means.
However, since the other choice of boundaries yields 2(1+e^2+e^4), its absence from the list leaves us no other, uh, choice.
I assumed circumscribed meant outside, but it is unclear indeed.
To approximate the integral using circumscribed rectangles, we divide the interval [0, 6] into 3 equal-width subintervals. The width of each subinterval would be (6 - 0) / 3 = 2.
The height of each rectangle is determined by evaluating the function at the right endpoint of each subinterval. In this case, the function is e^x, so the height of the first rectangle would be e^(2), of the second rectangle would be e^(4), and of the third rectangle would be e^(6).
Now, we calculate the area of each rectangle by multiplying the width and height. The area of the first rectangle is 2 * e^(2), the area of the second rectangle is 2 * e^(4), and the area of the third rectangle is 2 * e^(6).
To approximate the integral, we sum up the areas of all the rectangles.
Approximation = area of first rectangle + area of second rectangle + area of third rectangle
Approximation = 2 * e^(2) + 2 * e^(4) + 2 * e^(6)
Therefore, the answer is option a) 2e^2 + 4e^4 + 6e^6.