Calculus: Area between curves.?

Question

Sketch the region enclosed by the given curves: y=3x and y=3x^2. Decide whether to integrate with respect to x or y Then find the area of the region.

Answer 1

First thing you need to do is figure where the curves intersect. Do that by setting the two equations equal to each other.

3x = 3x^2
x=x^2
x = 0 and 1

You figure out the area below the curve for each curve over the interval from 0 to 1 and subtract. You may not know which is greater right off the bat, so you use the absolute value of the difference. (If you actually graph each curve, it will be pretty obvious in this case which curve has less area under it).

You find the area for each region by integrating:

int(3x) dx [1,0]= 3/2 x^2 [1,0] = 3/2
int(3x^2) dx [1,0] = x^3 [1,0] = 1

The difference, or the region bounded by the two curves, is:

3/2 - 1 = 1/2

Answer 2

find the points where the curves intersect, these will be the limits of integration.

x=0 makes y=0 in both equations so one of the intersections is 0,0
x=1 makes y =3 in both equations so the second intersection is 1,3

integrate with respect to x between x=0 and x=1

int [3x - 3x^2] = (3/2)x^2 -(3/3)x^3

putting in limits, area = (3/2)1 - (3/3)1^3 - 0 + 0
area = 3/2 - 3/3 =1/2

Answer 3

Okay, you have the curves y=3x and y=3x^2

First, as you are likely just starting into finding areas, I would draw a sketch like the question asks. You will find that the line y=3x is above the curve y=3x^2. Therefore, you need to find the two points of intersection:

3x=3x^2
So they intersect when x=1 and x=0 (thus the points are (0,0) and (1,3)).

Now, because y=3x is above the other curve, you subtract the lower curve from it and integrate that function with respect to x from 0 to 1 (the x-coordinates of the intersections; when you integrate with respect to some variable, the bounds are equal to that variable, which will come into play with multiple integrals):

Now integrate:

∫[0,1] (3x-3x^2) dx = (1.5x^2-x^3)|[0,1] = (1.5-1) - (0-0) = .5 sq. units

Answer 4

You get a "watermelon" slice turned sideways shape of a region between x=0 and x=1.
I'd integrate as S (0 to 1) 3x- 3x^2 dx.

Answer 5

y = 4x y = 2x² First, set both curves equivalent to discover their x-intercepts: 4x = 2x² via remark, x = 0. to boot: x = 4/2 = 2 hence, the bounds of integration are x = 0 to 2. A = ? 4x - 2x² dx A = [2x² - (2/3)x³] evaluated from x = 0 to 2 A = [2(2)² - (2/3)(2)³] - [2(0)² - (2/3)(0)³] = 8 - (2/3)(8) = 8/3 gadgets²

Answer 6

good one!
intergrate w/ respect to x.

find intersections so set them = to each other.
I got x=0 and x=1

area=to the integral from 0 to 1 of (3x)-(3x^2) dx
={ (3x^2/2) - (x^3) } from 0 to1 *hint:2nd FTOC
sub-in
so you should get 1/2
tell me i got it right ok! :)