Find the area bounded by the curves y=4x^2-25, y=0, x=-1, x=3.5. Ps the area is not 54 or 60 I already tried.
To find the area bounded by the curves, we can use definite integration. The area in this case would be equal to the definite integral of the upper curve minus the lower curve from the given x-limits.
First, let's find the x-values at which the two curves intersect. By setting the two equations equal to each other, we can solve for x:
4x^2 - 25 = 0
Simplifying, we have:
4x^2 = 25
x^2 = 25/4
x = ±√(25/4) = ±5/2
Since the interval of integration is defined between x = -1 and x = 3.5, we can see that the curve y = 4x^2 - 25 is the upper curve from x = -1 to x = 5/2, and the line y = 0 is the upper curve from x = 5/2 to x = 3.5.
To calculate the area, we need to split the integral into two parts, one for each region:
Area = ∫[from -1 to 5/2] (4x^2 - 25) dx + ∫[from 5/2 to 3.5] (0) dx
Evaluating the first integral:
∫[from -1 to 5/2] (4x^2 - 25) dx = [4 * (x^3)/3 - 25x] from -1 to 5/2
= [(4 * (5/2)^3)/3 - 25 * (5/2)] - [(4 * (-1)^3)/3 - 25 * (-1)]
= (500/6 - 125/2) - (-4/3 + 25)
= (250/3 - 125/2) - (-4/3 + 75/3)
= (250/3 - 125/2) - (71/3)
= 250/3 - (125/2 + 71/3)
Now, evaluating the second integral:
∫[from 5/2 to 3.5] (0) dx = 0 * x from 5/2 to 3.5 = 0
Therefore, the total area is:
Area = (250/3 - (125/2 + 71/3)) + 0 = 250/3 - (125/2 + 71/3)
To calculate the numerical value of the area, subtract the sum in the parentheses from 250/3.
Area ≈ 10.833 square units.
You posted this yesterday when you were Natalie
Here was my reply:
https://www.jiskha.com/display.cgi?id=1517283368
I had asked you to check my arithmetic, which you clearly did not do, since when I checked it just now it turned out to be 76 2/3 square units, not 60
(my arithmetic error was the 81/2 , should have been 343/6 )