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9.1 Determine the following integrals: 9.1.1 \( \int (10x + 6) \,dx \) 9.1.2 \( \int (x^2 + 2) - 2x^3 \,dx \) 9.2 The sketch below represents the shaded area bounded by the curve of the function defined by \( f(x) = \frac{3}{x} - 4; \: x > 0 \) and the x-axis between the points where \( x = 2 \) and \( x = 4 \) ![Figure](Figure_placeholder) Determine (showing ALL calculations) the shaded area bounded by the curve and the x-axis between the points where \( x = 2 \) and \( x = 4 \). - NSC Technical Mathematics - Question 9 - 2022 - Paper 1

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Question 9

9.1-Determine-the-following-integrals:--9.1.1-\(-\int-(10x-+-6)-\,dx-\)-9.1.2-\(-\int-(x^2-+-2)---2x^3-\,dx-\)--9.2-The-sketch-below-represents-the-shaded-area-bounded-by-the-curve-of-the-function-defined-by-\(-f(x)-=-\frac{3}{x}---4;-\:-x->-0-\)-and-the-x-axis-between-the-points-where-\(-x-=-2-\)-and-\(-x-=-4-\)--![Figure](Figure_placeholder)--Determine-(showing-ALL-calculations)-the-shaded-area-bounded-by-the-curve-and-the-x-axis-between-the-points-where-\(-x-=-2-\)-and-\(-x-=-4-\).-NSC Technical Mathematics-Question 9-2022-Paper 1.png

9.1 Determine the following integrals: 9.1.1 \( \int (10x + 6) \,dx \) 9.1.2 \( \int (x^2 + 2) - 2x^3 \,dx \) 9.2 The sketch below represents the shaded area bound... show full transcript

Worked Solution & Example Answer:9.1 Determine the following integrals: 9.1.1 \( \int (10x + 6) \,dx \) 9.1.2 \( \int (x^2 + 2) - 2x^3 \,dx \) 9.2 The sketch below represents the shaded area bounded by the curve of the function defined by \( f(x) = \frac{3}{x} - 4; \: x > 0 \) and the x-axis between the points where \( x = 2 \) and \( x = 4 \) ![Figure](Figure_placeholder) Determine (showing ALL calculations) the shaded area bounded by the curve and the x-axis between the points where \( x = 2 \) and \( x = 4 \). - NSC Technical Mathematics - Question 9 - 2022 - Paper 1

Step 1

9.1.1 \( \int (10x + 6) \,dx \)

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Answer

To find the integral ( \int (10x + 6) ,dx ):

  1. Apply the power rule for integration: [ \int x^n ,dx = \frac{x^{n+1}}{n+1} + C ]

    • For ( 10x ), this yields ( 10 \cdot \frac{x^2}{2} = 5x^2 ).
    • For ( 6 ), this yields ( 6x ).
  2. Combine results: [ 5x^2 + 6x + C ]

Step 2

9.1.2 \( \int (x^2 + 2) - 2x^3 \,dx \)

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Answer

To calculate ( \int (x^2 + 2 - 2x^3) ,dx ):

  1. Integrate each term separately:

    • For ( x^2 ), we have ( \frac{x^3}{3} ).
    • For ( 2 ), it becomes ( 2x ).
    • For ( -2x^3 ), it becomes ( -\frac{2}{4} x^4 = -\frac{1}{2} x^4 ).
  2. Combine the integrated terms: [ \frac{x^3}{3} + 2x - \frac{1}{2} x^4 + C ]

Step 3

9.2 Determine (showing ALL calculations) the shaded area bounded by the curve and the x-axis between the points where \( x = 2 \) and \( x = 4 \)

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Answer

To find the shaded area bounded by the curve and the x-axis from ( x = 2 ) to ( x = 4 ), we integrate:

  1. Set up the integral: [ A = \int_{2}^{4} \left( \frac{3}{x} - 4 \right) ,dx ]

  2. Calculate the definite integral:

    • First, integrate: [ \int \left( \frac{3}{x} - 4 \right) ,dx = 3 \ln |x| - 4x ]
    • Evaluate from 2 to 4: [ A = \left[ 3 \ln(4) - 4(4) \right] - \left[ 3 \ln(2) - 4(2) \right] ]
    • This simplifies to: [ A = (3 \ln(4) - 16) - (3 \ln(2) - 8) ]
    • Combine and simplify further to find approximate values: [ A \approx 5.92 \text{ square units} ]

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