Determine the following integrals: 9.1.1 $\int (x^{2} + 6x) \, dx$ 9.1.2 $\int (3x^{+} 1) \, dx$ The sketch below represents the area bounded by the function $g$ defined by: g(x) = 3x^{2} and the points where $x = k$ and $x = 4$ - NSC Technical Mathematics - Question 9 - 2021 - Paper 1

Similarly, we can compute this integral:

$\int (3x + 1) \, dx = \int 3x \, dx + \int 1 \, dx$

Calculating these integrals:

• The first term is: $\int 3x \, dx = \frac{3x^{2}}{2} + C$
• The second term is: $\int 1 \, dx = x + C$

Combining these gives: $\int (3x + 1) \, dx = \frac{3x^{2}}{2} + x + C.$

To find the area between the curve $g(x) = 3x^{2}$ from $x=k$ to $x=4$, we set up the integral:

$A = \int_{k}^{4} 3x^{2} \, dx$

Calculating this integral:

• First, compute: $\int 3x^{2} \, dx = x^{3} + C$

Thus, the definite integral becomes: $A = \left[ x^{3} \right]_{k}^{4} = 4^{3} - k^{3} = 64 - k^{3}$

Setting this equal to 56 gives: $64 - k^{3} = 56$

Solving for $k^{3}$: $k^{3} = 64 - 56 = 8 \Rightarrow k = 2.$