Final answer:
The area between the x-axis and the function f(x) = 5x^3 over the interval [-5, 3] is found using the definite integral. The final calculation provides the total area, which is 680 square units, representing the area enclosed by the curve and the x-axis over the given interval.
Step-by-step explanation:
Calculating Area Using Definite Integrals
To calculate the area between the x-axis and the function f(x) = 5x^3 over the interval [-5, 3], we should first determine if the function crosses the x-axis within this interval. Since the function is a cubic polynomial and there is no change in sign for f(x) over the given interval, it does not cross the x-axis. Thus, we can proceed with finding the definite integral directly.
The definite integral of f(x) over the interval [-5, 3] is performed as follows:
\[\int_{-5}^{3} 5x^3 dx\]
This integral computes the net area between the function and the x-axis over the interval. We calculate the antiderivative of f(x) first, and then evaluate it from the lower to upper limits of the interval:
\[F(x) = \frac{5}{4}x^4\]
\[Area = F(3) - F(-5)\]
\[Area = \frac{5}{4}(3^4) - \frac{5}{4}((-5)^4)\]
\[Area = \frac{5}{4}(81) - \frac{5}{4}(625)\]
\[Area = 101.25 - 781.25 = -680\]
Since we're seeking the magnitude of the area, we take the absolute value:
\[Area = | -680 | = 680\]
The total area under the curve from x = -5 to x = 3 is 680 square units.