9.1.1 Determine the following integrals: ∫ 3x² dx 9.1.2 ∫ (4 + 2^x) dx 9.1.3 ∫ rac{8x^3 - x^2}{2x} dx 9.2 The sketch below shows the shaded area bounded by function h defined by h(x) = -x² + 2x + 8 and the x-axis between the points where x = -2 and x = 4 - NSC Technical Mathematics - Question 9 - 2022 - Paper 1

We can split the integral into two parts:

$$\int (4 + 2^x) dx = \int 4 dx + \int 2^x dx.$$

Calculating each part gives:

1. For the first part,

$$\int 4 dx = 4x + C_1.$$

2. For the second part, using the integral of an exponential, we have:

$$\int 2^x dx = \frac{2^x}{\ln(2)} + C_2.$$

Combining the results, we arrive at:

$$\int (4 + 2^x) dx = 4x + \frac{2^x}{\ln(2)} + C.$$

We can simplify the integrand:

$$$$

Now we can integrate term by term:

$$\int (4x^2 - \frac{x}{2}) dx = \int 4x^2 dx - \int \frac{x}{2} dx.$$

Calculating each integral: 1.

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$$$$

Thus, the total integral becomes:

$$$$

To verify the learner's statement, we first need to calculate the area under the curve h from x = -2 to x = 4:

$$A = \int_{-2}^{4} h(x) dx = \int_{-2}^{4} (-x^2 + 2x + 8) dx.$$

Calculating, we get:

1. Find the antiderivative:

$$= \left[ -\frac{x^3}{3} + x^2 + 8x \right]_{-2}^{4}.$$

2. Evaluating at the limits:

$$= \left[ -\frac{4^3}{3} + 4^2 + 8\cdot4 \right] - \left[ -\frac{(-2)^3}{3} + (-2)^2 + 8 \cdot (-2) \right].$$

3. Calculation results in:

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We need to compute the entire area and check if it is indeed 20% of 36:

$$\text{Therefore, } 20\% \text{ of } 36 = 7.2.$$

A total area of approximately 25.93% confirms the learner's statement is NOT correct:

Conclusion: The learner's statement is NOT correct.