Bepaal die volgende integrale: 1 - NSC Technical Mathematics - Question 9 - 2023 - Paper 1

First, simplify the integrand:

$x^3 ( x^2 - 9 - x^6 ) = x^5 - 9x^3 - x^9$

Now, we can integrate each term separately:

$\int (x^5 - 9x^3 - x^9) dx = \frac{x^6}{6} - 9 \frac{x^4}{4} - \frac{x^{10}}{10} + C$

Thus, the complete result is:

$\frac{x^6}{6} - \frac{9 x^4}{4} - \frac{x^{10}}{10} + C$

The area under the curve can be calculated by integrating the function $f(x) = -x^2 + 2x + 3$ from point A to point B:

$A = \int_{-1}^{3} (-x^2 + 2x + 3) \, dx$

Now, calculate the integral:

1. Find the antiderivative:

   $= \left[-\frac{x^3}{3} + x^2 + 3x\right]_{-1}^{3}$

2. Evaluate at the limits:

   $= \left[-\frac{3^3}{3} + 3^2 + 9\right] - \left[-\frac{(-1)^3}{3} + (-1)^2 + 3(-1)\right]$

   At x = 3: $= -9 + 9 + 9 = 9$

   At x = -1: $= \frac{1}{3} + 1 - 3 = -\frac{5}{3}$

3. Combining these results gives:

   $= 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3}$

Thus, the total area under the curve between points A and B is approximately:

$\frac{32}{3} \text{ units}^2$