Area Bounded by a region

[hgaon001]

I have to find the area bounded by f(x)=2x/sqrt(9-x^2), g(x)=3/sqrt(9-x^2) and x=0

i know the area of a bounded region would be the integral of (f(x)-g(x))dx but i dont kno wat to do with the x=0... does that mean the interval changes?

 

[DrMike]

Yes, the x=0 tells you one of the bounds on the integral.

I'd suggest, with this type of question, *always* draw a picture, so you can see what region they are asking for.

 

[BigGlenntheHeavy]

\(\displaystyle A \ = \ \int_{0}^{\frac{3}{2}}\frac{(3-2x)dx}{\sqrt(9-x^{2})}\)

 

[fasteddie65]

Re: Area Bounded by a region (suggestion)

For the integral given in the previous post, I suggest you break it up into two separate integrals: ? 3/sqrt - ? 2x/sqrt. Then you will have to use a trig substitution on the first integral, and a u-substitution on the second.

 

[galactus]

For what it's worth, I tend to agree with pka's stance on this matter. If the student can set up the integral, use a calculator to solve it rather than go through all

that cumbersome trig sub and what not. Setting it up is the part that takes the human brain. After that, let a calculator do the grunt work. My opinion.

But, I must admit, sometimes I like to just evaluate integrals as a sort of puzzle. Especially, if they're tricky ones. As the \(\displaystyle \frac{sin(x)}{\sqrt{x}}\) in another post.