Find the area of the region bounded by
f(x)=5xsqrt(121−x2)
and the x-axis.
The area is = to
y = 5x√(121-x^2)
This is an odd function, so algebraically, the area is zero -- equal areas above and below the x-axis.
By symmetry, the geometric area is
a = 2∫[0,11] 5x√(121-x^2) dx
If you let u = 121-x^2, du = -2x dx, so
a = 2∫[121,0] -5/2 √u du
= -5 (2/3) u^(3/2) [121,0]
= 5 (2/3) 1331 = 13310/3
To find the area of the region bounded by the graph of the function f(x) and the x-axis, you can use integration.
First, let's determine the limits of integration. Since we are interested in the area bounded by the graph and the x-axis, we need to find the x-values where f(x) intersects the x-axis. In other words, we need to solve the equation f(x) = 0.
Given f(x) = 5x * sqrt(121 - x^2), we can set it equal to zero:
5x * sqrt(121 - x^2) = 0
Since the product of two factors is equal to zero, either 5x = 0 or sqrt(121 - x^2) = 0.
Solving 5x = 0, we find x = 0.
Solving sqrt(121 - x^2) = 0, we get x^2 = 121, which gives us x = ±11.
So, the limits of integration are x = -11 and x = 11.
To calculate the area, we can integrate the absolute value of f(x) from x = -11 to x = 11:
A = ∫[from -11 to 11] |f(x)| dx
Since f(x) is always positive for the given function, we can simplify it to:
A = ∫[from -11 to 11] f(x) dx
Now, we can integrate:
A = ∫[from -11 to 11] 5x * sqrt(121 - x^2) dx
To evaluate this integral, you can use integration techniques such as substitution or integration by parts.
Once you have evaluated the integral, you will have the area of the region bounded by f(x) and the x-axis.