Find the exact area below the curve y=x^3(4-x)and above the x-axis
the area is just
∫[0,4] x^3(4-x) dx
= x^4 - x^5/5 [0,4]
= 256 - 1024/5
= 256/5 or 51.25
To find the exact area below the given curve and above the x-axis, we need to integrate the function over the appropriate interval.
Here's how you can do it step by step:
1. Determine the interval over which the curve is above the x-axis. In this case, since the given curve is y = x^3(4-x), we need to find the x-values where y ≥ 0. Set the equation equal to zero and solve for x:
x^3(4 - x) = 0
The solutions are x = 0 (multiplicity 3) and x = 4 (multiplicity 1). So, the curve is above the x-axis on the interval [0, 4].
2. Integrate the function with respect to x over the interval [0, 4]. The integral represents the signed area between the curve and the x-axis. Use the following integral:
A = ∫[0, 4] x^3(4 - x) dx
3. Simplify the integrand and then evaluate the integral. Expand the expression and distribute the power of x:
A = ∫[0, 4] (4x^3 - x^4) dx
Now, integrate each term separately:
A = ∫[0, 4] 4x^3 dx - ∫[0, 4] x^4 dx
The integral of x^n is (x^(n+1))/(n+1), so we can use this to evaluate the integrals:
A = [(4x^4)/(4)] - [(x^5)/(5)] | [0, 4]
= [x^4 - (x^5)/(5)] | [0, 4]
Evaluating at the upper and lower limits:
A = (4^4 - (4^5)/(5)) - (0^4 - (0^5)/(5))
= 256 - (512/5)
4. Calculate the area:
A = 256 - (512/5) = 256 - 102.4 = 153.6 square units
Therefore, the exact area below the curve y = x^3(4-x) and above the x-axis is 153.6 square units.