Use the definite integral to find the area between the x-axis over the indicated interval.
f(x) = 36 - x^2; [-1,13]
So, what does be the area between the x-axis and f(x) equal? Thank you for any help! I'm really confused with this problem!
Do you mean algebraic (signed) area? If so, that is -686/3.
If you mean actual physical area, where area below the axis is also considered positive, then you have to break the area into the part above the axis [-1:6] and below the axis [6:13]. Doing it that way, we get 588.
588 is the correct answer? But, I don't understand how to get to that number. How did you calculate that? Sorry, that may be a lot to type out
588 is the correct answer!**
The graph crosses the x-axis at (6,0)
integrate 36-x^2 from -1:6 to get the positive area
integrate -(36-x^2) from 6:13 to add the area below the axis instead of subtracting it
∫[-1,6](36-x^2) + ∫[6,13](x^2-36)
= (36x - x^3/3)[-1:6] + (x^3/3 - 36x)[6,13]
539/3 + 1225/3 = 1764/3 = 588
To find the area between the x-axis and the curve defined by the function f(x) = 36 - x^2 over the interval [-1, 13], you can use the definite integral.
The definite integral represents the signed area between the curve and the x-axis over a given interval. In this case, since we are interested in the area above the x-axis, we can use the absolute value of the definite integral to find the positive area.
To calculate the area, you can set up the definite integral as follows:
Area = ∫[a,b] |f(x)| dx
In this case, a = -1 and b = 13, and f(x) = 36 - x^2.
Therefore, the integral to find the area becomes:
Area = ∫[-1,13] |36 - x^2| dx
To evaluate this integral, we can split it into two separate integrals based on the behavior of the function within the interval.
1. For x values where 36 - x^2 is positive:
The integral becomes ∫[-1,13] (36 - x^2) dx
2. For x values where 36 - x^2 is negative:
The integral becomes ∫[-1,13] -(36 - x^2) dx
Now we can evaluate each integral separately. We'll first find the common integral:
∫(36 - x^2) dx = 36x - (x^3/3) + C
For the first integral:
∫[-1,13] (36 - x^2) dx
= [36x - (x^3/3)] | from -1 to 13
= [(36*13 - (13^3)/3) - (36*(-1) - ((-1)^3)/3)]
Similarly, for the second integral:
∫[-1,13] -(36 - x^2) dx
= -[36x - (x^3/3)] | from -1 to 13
= -[(36*13 - (13^3)/3) - (36*(-1) - ((-1)^3)/3)]
Substituting the values into the expressions, you can evaluate both of these integrals to find the area between the x-axis and the curve over the interval [-1, 13].
Note: The absolute value is included only in the integral to ensure that the area is positive. The negative sign in the second integral is accounted for by subtracting the result of the second integral from the first integral.