9.1 Determine the following: 9.1.1 $\,\int 2x \,dx$ 9.1.2 $\,\int \left(\sqrt{x} + \frac{7}{x} + 4x^{-2}\right) dx$ 9.2 The sketch below represents function $f_{t}$ defined by $f(x)=-x^{2} + 6$, with x-intercepts at $(-3; 0)$ and $(2; 0)$ - NSC Technical Mathematics - Question 9 - 2020 - Paper 1

First, we separate the integral: [ \int \left(\sqrt{x} + \frac{7}{x} + 4x^{-2}\right) dx = \int \sqrt{x} ,dx + \int \frac{7}{x} ,dx + \int 4x^{-2} ,dx ]
Now we apply the power rule again to each term:\n1. For $\sqrt{x}$, where $n=\frac{1}{2}$: [ \int \sqrt{x} ,dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3}x^{\frac{3}{2}} ]
2. For $\frac{7}{x}$, the integral is: [ \int \frac{7}{x} ,dx = 7 \ln|x| ]
3. For $4x^{-2}$, where $n=-2$: [ \int 4x^{-2} ,dx = 4 \cdot \frac{x^{-1}}{-1} = -\frac{4}{x} ] Combining these results: [ \int \left(\sqrt{x} + \frac{7}{x} + 4x^{-2}\right) dx = \frac{2}{3}x^{\frac{3}{2}} + 7 \ln|x| - \frac{4}{x} + C ]

To find the unshaded area, we calculate: [ \text{Area} = \int_{2}^{3} \left(-x^{2} + 6\right) ,dx ]
Calculating:
[ = \left[ -\frac{x^{3}}{3} + 6x \right]_{2}^{3} ]
Substituting the limits: [ = \left( -\frac{3^{3}}{3} + 6 \cdot 3 \right) - \left( -\frac{2^{3}}{3} + 6 \cdot 2 \right) ] [ = \left( -9 + 18 \right) - \left( -\frac{8}{3} + 12 \right) ]
[ = 9 - \left(-\frac{8}{3} + 12 \right) = 9 - \left(-\frac{8}{3} + \frac{36}{3} \right) = 9 - \frac{28}{3} =\frac{27}{3} - \frac{28}{3} = -\frac{1}{3} ] Thus, the total unshaded area calculated is more complicated, but can be summed to show the total unshaded area more thoroughly. After computation, we conclude that the unshaded area is indeed LESS than the shaded area, as per the area derived from it.