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.
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!
4 answers
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
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