Find the area between the curves y= x^2/2 +2 and y = –x – 3 on the interval –4 ≤ x ≤ 4.
I got 40/3 is that correct?
64/6 + 8 + 20 - ( -64/6 + 8 -20 )
64/3 + 0 + 40
184/3
integral (x^2/2 + x + 5)dx
x^3 /6 + x^2/2 + 5 x at 4 minus at -4
64/6 + 8 + 20 - ( -64/6 + 8 -20 )
64/3 + 20 + 40
64/3 + 180/3 = 244/3 = 81 1/3
To find the area between the curves, you need to calculate the definite integral of the difference between the two functions over the given interval.
First, find the points of intersection between the two curves by setting them equal to each other and solving for x:
x^2/2 + 2 = -x - 3
Rearranging the equation, we get:
x^2/2 + x + 5 = 0
Using the quadratic formula, the solutions are:
x = (-1 ± √(1 - 4(1/2)(5)) / 2(1/2)
x = (-1 ± √(-19)) / 1
Since the solutions involve a complex number, there are no points of intersection between the two curves in the given interval.
Thus, we need to break the interval -4 ≤ x ≤ 4 into two parts: -4 ≤ x ≤ -1 and -1 ≤ x ≤ 4.
For the first part, -4 ≤ x ≤ -1, the curve y = -x - 3 is above y = x^2/2 + 2.
Therefore, the area under the curve y = -x - 3 and above y = x^2/2 + 2 in this interval can be calculated with the integral:
A1 = ∫(-4 to -1) [(-x - 3) - (x^2/2 + 2)] dx
Simplifying the integral, we get:
A1 = ∫(-4 to -1) [-x - 3 - x^2/2 - 2] dx
A1 = ∫(-4 to -1) [-x - x^2/2 - 5] dx
Using the power rule to integrate each term, we have:
A1 = [-(x^2/2) - (x^3/6) - 5x] evaluated from -4 to -1
Plugging in the limits of integration, we get:
A1 = [-((-1)^2/2) - ((-1)^3/6) - 5(-1)] - [(-(-4)^2/2) - ((-4)^3/6) - 5(-4)]
A1 = [(-1/2) - (-1/6) + 5] - [(8/2) - (64/6) + 20]
A1 = [-1/2 + 1/6 + 5] - [4 - 32/3 + 20]
A1 = [17/6] - [2/3]
A1 = (17 - 4) / 6
A1 = 13/6
For the second part, -1 ≤ x ≤ 4, the curve y = x^2/2 + 2 is above y = -x - 3.
So, the area under the curve y = x^2/2 + 2 and above y = -x - 3 in this interval can be calculated with the integral:
A2 = ∫(-1 to 4) [(x^2/2 + 2) - (-x - 3)] dx
Simplifying the integral, we get:
A2 = ∫(-1 to 4) [x^2/2 + 2 + x + 3] dx
A2 = ∫(-1 to 4) [x^2/2 + x + 5] dx
Using the power rule to integrate each term, we have:
A2 = [(x^3/6) + (x^2/2) + 5x] evaluated from -1 to 4
Plugging in the limits of integration, we get:
A2 = [(4^3/6) + (4^2/2) + 5(4)] - [((-1)^3/6) + ((-1)^2/2) + 5(-1)]
A2 = [64/6 + 16/2 + 20] - [(-1/6) + 1/2 - 5]
A2 = [32/3] - (7/6)
A2 = (32 - 7) / 3
A2 = 25/3
Finally, add the two areas together to get the total area between the curves:
Total area = A1 + A2 = (13/6) + (25/3) = 13/6 + 50/6 = 63/6 = 21/2
Therefore, the correct answer is 21/2, not 40/3.
To find the area between two curves, you need to integrate the difference between the two functions over the given interval.
In this case, we have two curves: y = (x^2/2) + 2 and y = -x - 3.
To find the area between these curves, we need to find the points where the two curves intersect. To do that, set the two functions equal to each other and solve for x:
(x^2/2) + 2 = -x - 3
Simplifying the equation, we get:
(x^2/2) + x + 5 = 0
Now, we can solve this quadratic equation. By factoring or using the quadratic formula, we find that the solutions are x = -4 and x = 2.
These are the x-values where the curves intersect. We need to integrate the difference between the two curves over the interval -4 ≤ x ≤ 4. We need to split the interval into two parts at x = 2 because the curves change.
For the left part of the interval (-4 ≤ x ≤ 2), the upper curve is y = (x^2/2) + 2 and the lower curve is y = -x - 3.
The integral to find the area of this part is:
∫[-4, 2] [(x^2/2) + 2 - (-x - 3)] dx
Simplifying the integral, we get:
∫[-4, 2] [(x^2/2) + 2 + x + 3] dx
= ∫[-4, 2] [x^2/2 + x + 5] dx
To find the antiderivative of each term, we get:
= [(x^3/6) + (x^2/2) + 5x] |[-4, 2]
Evaluating at the upper and lower limits, we have:
= [(2^3/6) + (2^2/2) + 5(2)] - [(-4^3/6) + (-4^2/2) + 5(-4)]
= (8/6 + 4 + 10) - (-64/6 + 8 + (-20))
= (16/12 + 48/12 + 120/12) - (-64/6 + 48/6 - 60/6)
= (184/12) - (-76/6)
= 92/6
= 46/3
So the area of the left part of the interval is 46/3.
For the right part of the interval (2 ≤ x ≤ 4), the upper curve is y = -x - 3 and the lower curve is y = (x^2/2) + 2.
Using the same process as above, we find that the area of this part is also 46/3.
Therefore, the total area between the curves on the interval -4 ≤ x ≤ 4 is the sum of these two areas:
Total Area = 46/3 + 46/3 = 92/3 = 30 2/3
So the correct answer is 92/3 or 30 2/3, not 40/3.