find the area between the x-axis and the graph of the given function over the given interval:
y = sqrt(9-x^2) over [-3,3]
you need to do integration from -3 to 3.
First you find the anti-derivative
when you find the anti-derivative you plug in -3 to the anti-derivative and then plug in 3 and find the difference of : f(-3)-f(3). Hope this helps :)
but how to you find the antideriv?
Paste sqrt(9-x^2) into
http://integrals.wolfram.com/index.jsp
This is an arcsin derivative.
Substitute x = 3 sin(t). Then t goes from - pi/2 to pi/2.
sqrt[9 - x^2] = 3 sqrt[1 - sin^2(t)] =
3 |cos(t)|
and dx = 3 cos(t) dt
So, you have to integrate 9 cos^2(t) from t = -pi/2 to pi/2 (note that in the interval cos(t) is positive so you can remoce the absolute value signs).
Since we are integrating over an entire period of the cos^2 the integral would be the same if you replace cos^2 by sin^2. By replacing cos^2 by cos^2 + sin^2 you thus obtain twice the value of the integral but since sin^2 + cos^2 = 1 that's 9 pi. So, the integral is 9/2 pi.
To find the area between the x-axis and the graph of the function y = sqrt(9 - x^2) over the interval [-3, 3], we need to use integration.
First, we need to find the antiderivative of the function sqrt(9 - x^2).
One way to find the antiderivative is to use an online integral calculator, such as the Wolfram Alpha website. You can simply paste the function into the calculator and it will provide you with the antiderivative. In this case, the antiderivative is arcsin(x/3).
Once you have the antiderivative, you can evaluate it at the endpoints of the interval [-3, 3]. Substitute -3 and 3 into the antiderivative function and subtract the values to find the difference: F(3) - F(-3).
In this case, the antiderivative arcsin(x/3) evaluated at 3 is arcsin(1) = pi/2 and evaluated at -3 is arcsin(-1) = -pi/2. So the difference is (pi/2) - (-pi/2) = pi.
Therefore, the area between the x-axis and the graph of y = sqrt(9 - x^2) over the interval [-3, 3] is pi square units.