Bepaal die volgende integrale: 9.1.1 $$\int \left(-\frac{6}{x}\right)dx$$ 9.1.2 $$\int (x-1)^{2}dx$$ 9.2 Die skets hieronder verteenwoordig die begrensde oppervlakklate van die kromme van die funksie wat deur $$f(x) = x^{3} + 3$$ gedefinieer word - NSC Technical Mathematics - Question 9 - 2018 - Paper 1

First, we expand the integrand:

$(x-1)^{2} = x^{2} - 2x + 1$

Now we can integrate each term separately:

$\int (x^{2} - 2x + 1) dx = \frac{x^{3}}{3} - x^{2} + x + C$

Thus, the final answer is:

$\frac{x^{3}}{3} - x^{2} + x + C$

To find the bounded area under the curve $f(x) = x^{3} + 3$ in the interval from $x = -2$ to $x = 1$, we compute the definite integral:

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

Calculating the integral:

$\int (x^{3} + 3) dx = \frac{x^{4}}{4} + 3x$

We evaluate this from $x = -2$ to $x = 1$:

$A = \left. \left( \frac{x^{4}}{4} + 3x \right) \right|_{-2}^{1} = \left( \frac{1^{4}}{4} + 3(1) \right) - \left( \frac{(-2)^{4}}{4} + 3(-2) \right)$

Calculating further, we find:

$A = \left( \frac{1}{4} + 3 \right) - \left( \frac{16}{4} - 6 \right) = \left( \frac{1}{4} + 3 \right) - \left( 4 - 6 \right) = \left( \frac{1}{4} + 3 \right) + 2$

Finally, simplifying yields:

$A = \frac{1}{4} + 5 = \frac{21}{4}$

Therefore, the bounded area is:

$\frac{21}{4}$