A long narrow piece of land gets flooded each year by a river. The flooded area is in the shape of the area under the curve y = 2.3 x^3 and above the x-axis, for 0 \le x \le 3.2. All the distances are in metres.
To determine the flooded area of the land, we need to find the area under the curve y = 2.3x^3 between the limits of x = 0 and x = 3.2.
To calculate the area under the curve, we can use integration. Integration essentially finds the antiderivative of a function, which gives us the area between the curve and the x-axis.
The general formula for finding the area under a curve between two limits is:
A = ∫[a to b] f(x) dx
In this case, the area A represents the flooded area, the function f(x) = 2.3x^3, and the limits of integration are a = 0 and b = 3.2.
To find the antiderivative of 2.3x^3, we use the power rule of integration:
∫ x^n dx = (1/n+1) * x^(n+1)
Applying this rule, we can integrate 2.3x^3:
∫ 2.3x^3 dx = (2.3/4) * x^4 + C
In this case, C represents the constant of integration, which we can ignore for now.
Now, we can evaluate the definite integral by plugging in the limits of integration:
A = [(2.3/4) * x^4] from 0 to 3.2
Substituting the limits:
A = [(2.3/4) * (3.2)^4] - [(2.3/4) * (0)^4]
Simplifying:
A = (2.3/4) * (3.2)^4
Calculating this expression will give you the flooded area in square meters.