Asked by Francesca
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!
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!
Answers
Answered by
Steve
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.
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.
Answered by
Francesca
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
Answered by
Francesca
588 is the correct answer!**
Answered by
Steve
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
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