Find the area lying above the x axis and below the parabolic curve y= 4x -x^2
a.8
b. 8 1/3
c. 10 2/3
d. 16
B or C not sure??
first, we find where the curve touches the x-axis, or simply get the roots,, to get the roots, equate y = 0:
0 = 4x - x^2
0 = x(4-x)
x = 0 and x = 4
this will be our boundary conditions for integration,, we'll use integration since area under the curve:
integral (4x - x^2) = 2x^2 - (1/3)X^3
at x = 0, the integral becomes
2*0^2 - (1/3)*0^3 = 0
at x = 4 the integral becomes
2*(4^2) - (1/3)*4^3
32 - 64/3
96/3 - 64/3
32/3 = 10 2/3
thus integral from 0 to 4 = 10 2/3 - 0 = 10 2/3
hope this helps~ :)
To find the area between a curve and the x-axis, you need to integrate the function that represents the curve with respect to x. In this case, you want to find the area above the x-axis and below the parabolic curve y = 4x - x^2.
First, you need to find the x-values where the curve intersects the x-axis. To do this, set y equal to zero and solve for x:
0 = 4x - x^2
Rearrange the equation:
x^2 - 4x = 0
Factor out an x:
x(x - 4) = 0
Set each factor equal to zero and solve for x:
x = 0 or x - 4 = 0
So the curve intersects the x-axis at x = 0 and x = 4.
Next, you need to determine the limits of integration. Since the area you want to find lies above the x-axis, the lower limit will be x = 0. And since the area lies below the parabolic curve, the upper limit will be x = 4.
Now, integrate the function y = 4x - x^2 with respect to x within the given limits:
∫[0 to 4] (4x - x^2) dx
To simplify the integration, split the integral into two parts:
∫[0 to 4] 4x dx - ∫[0 to 4] x^2 dx
Evaluate each integral:
∫[0 to 4] 4x dx = 2x^2 |[0 to 4] = 2(4)^2 - 2(0)^2 = 32
∫[0 to 4] x^2 dx = (1/3)x^3 |[0 to 4] = (1/3)(4)^3 - (1/3)(0)^3 = 64/3
Finally, subtract the second integral from the first:
32 - (64/3) = 32 - 21.33 = 10.67
Therefore, the area lying above the x-axis and below the parabolic curve y = 4x - x^2 is approximately 10.67.
Looking at the answer choices, the closest option is c. 10 2/3.