Integrals? Can you help....?

Question

Thanks to those of you who helped with my Chain Rule problem. Thats what I got but the buck had a different answer. Anywho, since I'm reviewing, this is another one I'm having trouble with.

Find the area of the between the curve: y=x^3, and the x-axis,from x=-2 to x=3.

Answer 1

y = x³ has a zero between -2 and 3 namely at x = 0
So you need to do the integral in two parts otherwise you will get a vector addition (as below the axis the area is negative)

Area = |Integral between x = -2 and x = 0) (x³)dx| + |Integral between x = 0 and x = 3) (x³)dx
= | [¼x^4] between x = -2 and x = 0 | + | [¼x^4] between x = 0 and x = 3 |
= ¼( | [0^4 - (-2)^4] | + | [3^4 - 0^4] |
=¼[16 + 81]
=97/4
(= 24¼)

Answer 2

y=x^3 is below the x-axis between -2 and 0, and above the x-axis between 0 and 3. Therefore, the total area of the enclosed region is given by the integrals

integral from -2 to 0 (0 - x^3) dx + integral from 0 to 3 (x^3-0) dx.

In each integral, your integrand is given by the y-coordinate of the top curve minus the y-coordinate of the bottom curve. In general, when computing the area between two curves, it's important to take into account the intersections of the curves to keep track of which is top and which is bottom.

Evaluating these integrals gives you
-x^4/4 from -2 to 0 + x^4/4 from 0 to 3
= 4 + 81/4
= 24 1/4

Answer 3

integrate y=x^3 from 0 to 3
and y=x^3 from -2 to 0
(x^4/4] from 0 to 3 - (x^4/4] from -2 to 0
(81/4-0) - (0-16/4) = 24.25

you do them both seperately because from -2 to 0 the curve is below the x-axis so it is going to be negative and you have to add the modulus of it.

Answer 4

x^3
=3x^2

then the ares is A=3(3)^2-3(-2)^2
=27-12
A=15