Pre-Calculus help?

Question

Use a definite integral to find the area between f(x)=1-x^2 and the x-axis over the interval [-3,0].

I have the answer, but I don't know how to get it, which is what I need to understand.

Answer 1

Definite integrals in pre-calc? OK...

∫(1 - x^2)dx - [-3, 0]

[x - (1/3)x^3] - [-3, 0]

[0 - (1/3)0^3] - [(-3) - (1/3)(-3)^3]

0 - [-3 - (1/3)(-27)]

- [-3 - (-9)]

-6

It's either positive or negative 6 depending on which order the upper/lower limits of integration were, but since your parabola has a negative concavity, -6 sounds right - area UNDER the x-axis would be negative, although there's no problem in saying that the area (which normally isn't a vector) is 6.

Hope that's what you needed.

Answer 2

well first you need the definite integral to figure it out

x-((1/3)x^3)

so now just use the interval that you have been given

(-3)-((1/3)(-3)^3)
-3-((1/3)(27))
-3-(-9)
-3+9
so that gives you six and since plugging in zero would simple give you zero you would get 6-0 which is still 6 and you have your area.

Answer 3

integral 1-x^2 from -3 to 0

=x-(x^3)/3 from -3 to 0
=0-(-3+27/3)
=-(-3+9)
=-6

answer: 6

Answer 4

This is a little ambiguous because the function crosses the x-axis in that region. If you need the total area, you need to find where it crosses, then find two separate integrals and add them. If you need the net area, you just need one integral.

Answer 5

0~-3 (1-x^2)

*split it up into 2 different intergrals:
0~-3 (1) -- 0~-3(x^2)
*Take the seperate intergrals:
(1x^0)==> 1x^1
(x^2)===> 1/3 x^3
*Now plug in the numbers
(1(0)^1--1/3(0)^3) -- (1(-3)^1--1/3(-3)^3)
Vuala!

Answer 6

A = ∫1 - x² dx between lims. - 3 and 0
A = [ x - x³ / 3 ] between -3 and 0
A = - 3 + 9
A = 6