9.1 Determine the following integrals: 9.1.1 \( \int \frac{6}{x} \; dx \) 9.1.2 \( \int (x-1)^{2} \; dx \) 9.2 The sketch below represents the bounded area of the curve defined by \( f(x) = x^{2} + 3 \) Determine the shaded area bounded by the curve and the \( x \)-axis between the points where \( x = -2 \) and \( x = 1 \). - NSC Technical Mathematics - Question 9 - 2018 - Paper 1

First, expand the integrand:

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

Now we can integrate term by term:

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

To find the area bounded by the curve ( f(x) = x^{2} + 3 ) and the x-axis, we first determine the points where the curve intersects the x-axis:

Set ( f(x) = 0 ):

$x^{2} + 3 = 0 \Rightarrow x^{2} = -3$

Since there are no real roots, the curve does not intersect the x-axis. We compute the area from ( x = -2 ) to ( x = 1 ) directly:

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

Calculating the integral:

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

Evaluating at the boundaries:

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

$= \left( \frac{1}{3} + 3 \right) - \left( -\frac{8}{3} - 6 \right)$

This will yield the total area after proper calculation.