Find the area of the region bounded by y=x^2 and y=9/(x^2+8)?

Question

I know how to find the intersection between two curves, but i cannot find the integral of y=9/(x^2+8)!

Give me the answer as soon as possible thanks!!!!

Answer 1

http://integrals.wolfram.com/index.jsp?e...

Answer 2

u^2+8u-9=0,

Where u = x^2

The quadratic equation 1x▓+ 8x + -9 = 1*(x-1)(x+9), x=-4 - Sqrt[100]/2 , x=-4 + Sqrt[100]/2

x^2 = 1

Its over [-1,1]

x^2 + 8x + -9 = 1*(x-1)(x+9), x=-4 - Sqrt[100]/2 , x=-4 + Sqrt[100]/2

Nevermind all that, it's sloppy I will just give you the integral of y=9/(x^2+8).

ANSWER:

º9dx/(8+x^2)

x = sqrt(8)tan(q)

dx = sqrt(8)sec^2(q)dq

º9sqrt(8)sec^2(q)/8(1+tan(q)^2)

º9sqrt(8)sec^2(q)dq/8(1+tan(q)^2)

º9sqrt(8)sec^2(q)dq/(8(1+tan(q)^2))

º9sqrt(8)sec^2(q)dq/(8(sec^2(q)))

º9sqrt(8)dq/(8)

º9sqrt(2)dq/4

9sqrt(2)/4ºdq

9sqrt(2)/4 * (q)

q = arctan(x/sqrt(8))

9sqrt(2)/4 * arctan(x/sqrt(8)) + C is the integral, indefinite. If it's a definite integral you can drop the C and plug in the bounds.

I should get some thumbs up for this I spent a lot of time. Sorry my math is a bit rusty.

Answer 3

By graphing, the points of intersections are (-1,1) and (1,1)

You need to integrate 9/(x^2+8)-x^2 from -1 to 1.

8= [2 sqrt(2)]^2

∫ 9 / (x^2 + [2 sqrt(2)]^2 dx - ∫ x^2 dx from  [-1,1]

= {9/ 2 sqrt(2)} arctan( x / 2sqrt(2)) - x^3/3 , limits [-1,1]

You can simplify the integral.

We have used the formula:

Integral of dx / x^2+8 = (1/sqrt(8)) arctan(x/sqrt(8))