Find the centroid of the area bounded by the parabola y = 4 - x^2 and the x-axis.
A. (0,1.6)
B. (0,1.7)
C. (0,1.8)
D. (0,1.9)
Find the average value of the function i=15(1-3^-1/2t) from t=0 to t=4.
A. 7.5(1+e^-2)
B. 7.5(2+e^-2)
C. 7.5(2-e^-2)
D. 7.5(3-e^-2)
I get (A)
Assuming the typo 3 means e, I get (B)
To find the centroid of the area bounded by a curve and the x-axis, you can use calculus to determine the coordinates of the centroid.
1. Start by setting up the integral to find the centroid. The formula for the y-coordinate of the centroid, denoted as yc, is given by:
yc = 1/A ∫[a,b] (x*f(x)) dx
Where A is the area bounded by the curve, f(x) is the equation of the curve, and [a,b] represents the interval over which the curve is bounded.
2. In this case, the equation of the curve is y = 4 - x^2 and it is bounded by the x-axis. To find the area A, we need to integrate the equation of the curve and set it equal to zero (as it touches the x-axis).
∫[a,b] (4 - x^2) dx = 0
3. Solving the integral, we find:
4x - (x^3)/3 = 0
x(4 - x^2)/3 = 0
x = 0 or x = ±2
So, the curve intersects the x-axis at x = 0 and x = ±2. We only need to consider the positive x-values as the curve is symmetric.
4. Now, we can calculate the area A. The formula for the area bounded by the curve is given by:
A = ∫[a,b] |f(x)| dx
In this case, the area A is:
A = ∫[0,2] (4 - x^2) dx
A = [4x - (x^3)/3] evaluated from 0 to 2
A = (8 - 8/3) - (0 - 0) = 16/3
So, the area bounded by the curve is 16/3.
5. Now, we can find the y-coordinate of the centroid, yc, by evaluating the integral mentioned in step 1:
yc = 1/A ∫[0,2] (x*(4 - x^2)) dx
yc = 1/(16/3) ∫[0,2] (4x - x^3) dx
yc = 3/16 [2x^2 - (x^4)/4] evaluated from 0 to 2
yc = 3/16 [(2*2^2 - (2^4)/4) - (2*0^2 - (0^4)/4)]
yc = 3/16 (8 - 4) = 3/16 * 4 = 3/4 = 0.75
Therefore, the y-coordinate of the centroid is 0.75.
6. Finally, the coordinates of the centroid are (0, 0.75).
So, the correct answer for the centroid of the area bounded by the parabola y = 4 - x^2 and the x-axis is not provided in the options given.