# Thread: Check: Area of region of plane between 2 graphs

1. ## Check: Area of region of plane between 2 graphs

y = 9 - x^2 and y = 1 + 2x

9 - x^2 = 1 + 2x
x^2 - 2x - 8 = 0
(x-4)(x+2) = 0
x = 4 & -2

Interval from -2 to 4: [(9-x^2) - (1+2x)] dx = -x^3 / 3 - x^2 = F(4) - F(-2)

F(4) = -(4)^3/3 - (4)^2 = 64/3 - 16 = 16/3

F(-2) = -(-2)^3 / 3 - (-2)^2 = -8/3 - 12/3 = -20/3

So F(4) - F(-2) = 16/3 - 20/3 = -4/3

ja oder nein?

2. Originally Posted by shepherdm1270
y = 9 - x^2 and y = 1 + 2x

9 - x^2 = 1 + 2x
x^2 - 2x - 8 = 0 Mr F says: It should be x^2 + 2x - 8 = 0. Here's a tip: Always substitute your answers back into the equations as a check. Ideally you will also draw the curves and shade the required region. I haven't checked past your solutions for the intersection points.

(x-4)(x+2) = 0
x = 4 & -2

Interval from -2 to 4: [(9-x^2) - (1+2x)] dx = -x^3 / 3 - x^2 = F(4) - F(-2)

F(4) = -(4)^3/3 - (4)^2 = 64/3 - 16 = 16/3

F(-2) = -(-2)^3 / 3 - (-2)^2 = -8/3 - 12/3 = -20/3

So F(4) - F(-2) = 16/3 - 20/3 = -4/3

ja oder nein?
..

3. Firstly, you know you made a mistake when you get a negative area.

You have to know the limits therefore you must calculate the intersections betwen the two curves.

9-x^2=1+2x

From this quatdration equation you can deduce using the Wiet's rule for quadratic formulae that intersections are x= -4 and x=2 which are now the limits.

Then you integrate each of the equations with respect to x from -4 to 2.

When you get the two results for integrals you must (!) take the absolute value for the difference between them. Then you will obtain positive value and true anwser.

4. Firstly, you know you made a mistake when you get a negative area.

You have to know the limits therefore you must calculate the intersections betwen the two curves.

9-x^2=1+2x

From this quatdration equation you can deduce using the Wiet's rule for quadratic formulae that intersections are x= -4 and x=2 which are now the limits.

Then you integrate each of the equations with respect to x from -4 to 2.

When you get the two results for integrals you must (!) take the absolute value for the difference between them. Then you will obtain positive value and true anwser.

5. Originally Posted by cigster
Firstly, you know you made a mistake when you get a negative area.
Hmm, what about $\displaystyle \displaystyle \int_{-\frac{\pi}{2}}^{0} \sin x \, dx = \displaystyle - \cos x |_{-\frac{\pi}{2}}^{0} = -1 + 0 = -1$?

6. Originally Posted by Isomorphism
Hmm, what about $\displaystyle \displaystyle \int_{-\frac{\pi}{2}}^{0} \sin x \, dx = \displaystyle - \cos x |_{-\frac{\pi}{2}}^{0} = -1 + 0 = -1$?
The area between two curves is always positive. It's found by finding int ((upper curve) - (lower curve).

The example you've given is a definite integral - it does not give the area between two curves.

The area between the sine curve and the x-axis from x = -pi/2 to x = 0 is given by int (0) - (sin x) ......