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Determine the following integrals: 9.1.1 \( \int (x^{2} + \frac{1}{x}) \, dx \) 9.1.2 \( \int (x^{3} - 5x^{4}) \, dx \) 9.2 The sketch below represents the shaded bounded area of the curve of the function defined by \( f(x) = 2x^{3} - 4 \) - NSC Technical Mathematics - Question 9 - 2021 - Paper 1
Question 9
Determine the following integrals: 9.1.1 \( \int (x^{2} + \frac{1}{x}) \, dx \) 9.1.2 \( \int (x^{3} - 5x^{4}) \, dx \) 9.2 The sketch below represents the shad... show full transcript
Worked Solution & Example Answer:Determine the following integrals: 9.1.1 \( \int (x^{2} + \frac{1}{x}) \, dx \) 9.1.2 \( \int (x^{3} - 5x^{4}) \, dx \) 9.2 The sketch below represents the shaded bounded area of the curve of the function defined by \( f(x) = 2x^{3} - 4 \) - NSC Technical Mathematics - Question 9 - 2021 - Paper 1
Step 1
9.1.1 \( \int (x^{2} + \frac{1}{x}) \, dx \)
Answer
To solve this integral, we can break it into two parts:
- ( \int x^{2} , dx = \frac{x^{3}}{3} + C_{1} )
- ( \int \frac{1}{x} , dx = \ln |x| + C_{2} )
Combining these, we have:
[ \int (x^{2} + \frac{1}{x}) , dx = \frac{x^{3}}{3} + \ln |x| + C ]
Step 2
9.1.2 \( \int (x^{3} - 5x^{4}) \, dx \)
Answer
This integral can also be solved by splitting it:
- ( \int x^{3} , dx = \frac{x^{4}}{4} + C_{1} )
- ( \int -5x^{4} , dx = -5 \cdot \frac{x^{5}}{5} + C_{2} = -x^{5} + C_{2} )
Thus, combining these:
[ \int (x^{3} - 5x^{4}) , dx = \frac{x^{4}}{4} - x^{5} + C ]
Step 3
9.2 Determine (showing ALL calculations) the shaded area bounded by the curve and the x-axis between the points where \( x = -1 \) and \( x = 2 \).
Answer
To find the area under the curve ( f(x) = 2x^{3} - 4 ) from ( x = -1 ) to ( x = 2 ), we first compute the definite integral:
[ A = \int_{-1}^{2} (2x^{3} - 4) , dx ]
Calculating this integral, we have:
[ = \left[ \frac{2x^{4}}{4} - 4x \right]{-1}^{2} ] [ = \left[ \frac{x^{4}}{2} - 4x \right]{-1}^{2} ] [ = \left( \frac{(2)^{4}}{2} - 4(2) \right) - \left( \frac{(-1)^{4}}{2} - 4(-1) \right) ] [ = \left( \frac{16}{2} - 8 \right) - \left( \frac{1}{2} + 4 \right) ] [ = (8 - 8) - (0.5 + 4) = 0 - 4.5 = -4.5 ]
Since we are interested in area, we take the absolute value:
[ A = 4.5 \text{ square units} ]