9.1 Determine the following integrals: 9.1.1 \( \int \frac{3}{x} \, dx \) 9.1.2 \( \int \left( \frac{3x}{x^2} + \sqrt{x} \right) \, dx \) 9.2 The sketch below shows function \( f \) defined by \( f(x) = x^2 - 5x \) The two shaded areas represented are: - \( A_1 = \) area bounded by function \( f \), the x-axis and the ordinates \( x = 0 \) and \( x = 3 \) - \( A_2 = \) area bounded by function \( f \), the x-axis and the ordinates \( x = 5 \) and \( x = 7 \) Determine (showing ALL calculations) by how much \( A_1 \) is greater than \( A_2 \) - NSC Technical Mathematics - Question 9 - 2023 - Paper 1

First, we simplify the integrand:

$\frac{3x}{x^2} = \frac{3}{x} \text{ and } \sqrt{x} = x^{1/2}$

Thus,

$\int \left( \frac{3}{x} + x^{1/2} \right) \, dx = \int \frac{3}{x} \, dx + \int x^{1/2} \, dx$

Now, integrating term by term:

• The integral of ( \frac{3}{x} , dx ) is ( 3 \ln |x| + C_1 ).
• The integral of ( x^{1/2} , dx ) is ( \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} + C_2 ).

Combining these results, we have:

$= 3 \ln |x| + \frac{2}{3} x^{3/2} + C$.

To find the areas ( A_1 ) and ( A_2 ), we calculate:

Area ( A_1 ):
( A_1 = \int_0^3 (x^2 - 5x) , dx )
Calculating the definite integral:

$= \left[ \frac{x^3}{3} - \frac{5x^2}{2} \right]_0^3 = \left( \frac{3^3}{3} - \frac{5(3^2)}{2} \right) - \left( 0 - 0 \right)$

$= \left( 9 - \frac{45}{2} \right) = 9 - 22.5 = -13.5 \text{ (area is positive)}$
Thus, ( A_1 = 13.5 \text{ units}^2 ).

Area ( A_2 ):
( A_2 = \int_5^7 (x^2 - 5x) , dx )
Calculating the definite integral:

$= \left[ \frac{x^3}{3} - \frac{5x^2}{2} \right]_5^7 = \left( \frac{7^3}{3} - \frac{5(7^2)}{2} \right) - \left( \frac{5^3}{3} - \frac{5(5^2)}{2} \right)$

Calculating these values:

$= \left( \frac{343}{3} - \frac{245}{2} \right) - \left( \frac{125}{3} - \frac{125}{2} \right)$

Calculating each term gives: $= 12.67 \text{ units}^2$

Finding the difference:
( A_1 - A_2 = 13.5 - 12.67 = 0.83 \text{ units}^2 )

Thus, ( A_1 ) is greater than ( A_2 ) by ( 0.83 \text{ units}^2 ).