area under the graph??!

[redragtop12]

Find the area under the graph to the x axis from -2 to 2 for

f(x) = (4 - x2)(1/2)

I know that I need to find the anti-deritative first...so i can plus in my numbers and subtract
but i am confused on how to do the anti derivative for that..

 

[royhaas]

Try the trig substitution \(\displaystyle x=2\sin(\theta)\).

 

[BigGlenntheHeavy]

\(\displaystyle f(x) = \int_{-2}^{2}\sqrt(4-x^{2})dx, = 2\int_{0}^{2}\sqrt(4-x^{2})dx. \ Let \ x = 2sin\theta, \ then \ dx = 2cos \theta d\theta.\)

\(\displaystyle Now, \ 2\int_{0}^{\frac{\pi}{2}}\sqrt(4-4sin^{2}\theta)(2cos\theta)d\theta.\)

\(\displaystyle Ergo, \ 8 \int_{0}^{\frac{\pi}{2}}cos^{2}\theta d\theta = 4 \int_{0}^{\frac{\pi}{2}} \ [1+cos(2\theta)]d\theta.\)

\(\displaystyle Therefore, \ 4 \int_{0}^{\frac{\pi}{2}} \ [1+cos(2\theta)]d\theta = 4[\theta+\frac{sin(2\theta)}{2}]_{0}^{\frac{\pi}{2}\)

\(\displaystyle = 4[\theta + sin(\theta)cos(\theta)]_{0}^{\frac{\pi}{2}} = 2\pi.\)

 

[galactus]

Why don't you just use ^ (SHIFT 6) for power. Wouldn't that make sense?. The way it is written looks like you are multiplying (4-x2) by 1/2.

That is \(\displaystyle (4-2x)(1/2)=2-x\)

But you mean \(\displaystyle \sqrt{4-x^{2}}\)

See what I mean?.

 

[redragtop12]

Yikes, you are correct I meant
(4-2x)^ (1/2)

 

[mmm4444bot]

… I meant (4-2x)^(1/2)

Are you sure? Your first version shows the base as 4 - x2.

The base is now 4 - 2*x versus 4 - x^2. :?