estimate as a trapezoid:
height at x =2 is 1/8
height at x=5 = 1/125
average height = (1/8+1/125)/2 = 133/2000
appr area = (3)(133/2000)
= 399/2000
= appr .2
actual area
= ∫1/x^3 dx from 2 to 5
= ( - 1/(2x^2) ) from 2 to 5
= (-1/2)( 1/25 - 1/4)
= 21/200
= .105
Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area.
y = x^−3, 2 ≤ x ≤ 5
2 answers
You can see a nice graph with grid lines here:
http://rechneronline.de/function-graphs/
You may have to adjust the y scale some so it fits. You can use that to estimate the area. I'm sure you can then evaluate the integral.
http://rechneronline.de/function-graphs/
You may have to adjust the y scale some so it fits. You can use that to estimate the area. I'm sure you can then evaluate the integral.