Find the area between each curve and the x axis for the given interval.
y=16x-x^3 from x = 0 to x = 4
Thanks ahead of time.
Please let us see how you have tried to do this integral.
To find the area between the curve and the x-axis for a given interval, you can use definite integration. Here's how you can find the area for the given curve and interval:
1. Start by finding the points where the curve intersects the x-axis. These points are where the y-coordinate is equal to zero. Let's find these points for the given curve y = 16x - x^3:
16x - x^3 = 0
x(16 - x^2) = 0
So, either x = 0 or 16 - x^2 = 0.
For x = 0, the y-coordinate is also zero. So, one of the points is (0, 0).
For 16 - x^2 = 0, solving for x:
x^2 = 16
x = ±4
So, the other two points where the curve intersects the x-axis are (4, 0) and (-4, 0).
2. The area between the curve and the x-axis for the given interval [0, 4] can be found by taking the integral of the absolute value of the curve from x = 0 to x = 4:
Area = ∫[0, 4] |16x - x^3| dx
However, since the curve is above the x-axis for the given interval, we can simplify the absolute value to just the curve itself:
Area = ∫[0, 4] (16x - x^3) dx
3. Now, we can calculate the integral:
Area = ∫[0, 4] (16x - x^3) dx
= [8x^2 - (x^4)/4] evaluated from 0 to 4
= [(8*4^2 - (4^4)/4)] - [(8*0^2 - (0^4)/4)]
= [128 - 16] - [0 - 0]
= 112
So, the area between the curve y = 16x - x^3 and the x-axis for the interval [0, 4] is 112 square units.