Asked by rosie

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 ≤ 6
Answered by rosie
a small typo,
y = x^-3, 2 ≤ x ≤ 6
Answered by Anonymous
integral y dx = x^2 - 3 x
at x = 6 - at x = 2

at x = 6
36 - 3*6 = 18
at x = 2
4 - 6 = -2
so
18 - -2 = 18 + 2 = 20
Answered by Anonymous
that is a big typo
y = x^-3, 2 ≤ x ≤ 6

integral of x^-3 dx = (-1/2) x^-2
so
(-1/2)(1/36 - 1/4) = (-1/2) (1/36 - 9/36)= (-1/2)(-8/36) = 4/36 = 1/9
Answered by oobleck
Somehow I think you meant y = x^2 - 3
since that curve lies entirely above the x-axis on the interval [2,6]
If you just want an estimate, consider the trapezoid with area
(1+33)/2 * 4 = 68

The actual area is
1/3 x^3 - 3x [2,6]
= (216/3 - 18) - (8/3 - 6) = 57 1/3
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