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
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
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
a small typo,
y = x^-3, 2 ≤ x ≤ 6
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
To estimate the area of the region beneath the curve, we can use a graph. Here's how you can do it:
1. Draw the x and y-axis on a Cartesian coordinate system.
2. Plot the curve y = x-3 on this graph.
3. Shade the region between the curve and the x-axis for the interval 2 ≤ x ≤ 6.
Now, to find the exact area of the region, we will use integration. The area is given by the definite integral of the function y = x-3 over the interval [2, 6]. Here's how you can do it:
1. Calculate the indefinite integral of the function y = x-3.
∫(x-3) dx = (x^2/2) - (3x) + C, where C is the constant of integration.
2. Evaluate the definite integral over the interval [2, 6].
∫[2, 6] (x-3) dx = [(x^2/2) - (3x)] evaluated between 2 and 6.
3. Plug in the upper limit (x = 6) into the integral expression:
[(6^2/2) - (3*6)].
4. Plug in the lower limit (x = 2) into the integral expression:
[(2^2/2) - (3*2)].
5. Subtract the result of the lower limit evaluation from the upper limit evaluation:
[(6^2/2) - (3*6)] - [(2^2/2) - (3*2)].
6. Simplify the expression:
[(36/2) - 18] - [(4/2) - 6] = [18 - 18] - [2 - 6] = 0 - (-4) = 4.
The exact area of the region beneath the curve y = x-3, for the interval 2 ≤ x ≤ 6, is 4 square units.