Graph the integrands and use areas to evaluate integrals
(integrate(3&-3)) root(9-x^2) dx
fix the 3&-3 typo, and try again. You will be doing a trig substitution. Why doncha also show some work, so we can see where you get stuck?
To graph the integrand, we need to first identify the function and its domain. In this case, the integrand is the function f(x) = √(9 - x^2), and its domain is all values of x that satisfy the inequality 9 - x^2 ≥ 0.
Let's solve this inequality to find the domain:
9 - x^2 ≥ 0
Rearranging, we get:
x^2 ≤ 9
Taking the square root of both sides, we have two cases:
Case 1:
x ≤ 3
Case 2:
x ≥ -3
Combining both cases, the domain of f(x) is -3 ≤ x ≤ 3.
Now, let's graph the function f(x) = √(9 - x^2):
To plot the points, we can start with x = -3 and substitute different values of x within the domain into the equation to find the corresponding y-values. However, this process can be time-consuming.
Alternatively, we can use technology such as graphing calculators or online graphing tools to quickly plot the graph. Here is a graph of f(x) = √(9 - x^2):
[Graph Image]
Now, to evaluate the integral ∫√(9 - x^2) dx from -3 to 3, we can use areas.
The integral ∫√(9 - x^2) dx represents the area between the curve f(x) = √(9 - x^2) and the x-axis, bounded by the limits of integration (-3 and 3).
In this case, we can observe that the function is symmetric about the y-axis, meaning that the areas on both sides of the y-axis cancel each other out. Therefore, we can find the area in one of the regions and double it to get the total area.
Since the function is the upper half of a circle with radius 3, the area is π * (3^2) / 2 = 9π / 2.
Finally, to evaluate the integral over the given limits:
∫(from -3 to 3) √(9 - x^2) dx = 2 * (9π / 2) = 9π