9.1 Determine the following: 9.1.1 $ \int 2x \: dx $ 9.1.2 $ \int \left( \sqrt{x} + \frac{7}{x} + 4x^{-2} \right) \: dx $ 9.2 The sketch below represents function $ f $ defined by $ f(x) = -x^2 + 6 $, with $ x $-intercepts at (-3; 0) and (2; 0). The shaded area, bounded by the curve and the $ x $-axis between the points $ x = -1 $ and $ x = 1 $, is shown in the sketch. The area of the shaded part is $ \frac{34}{3} $ square units. Determine whether the unshaded area, bounded by the curve and the $ x $-axis between the points $ x = 3 $ and $ x = 2 $, is less than the shaded area.

NSC Technical Mathematics Differential Calculus and Integration: 9.1 Determine the following: 9.

9.1 Determine the following: 9.1.1 $ \int 2x \: dx $ 9.1.2 $ \int \left( \sqrt{x} + \frac{7}{x} + 4x^{-2} \right) \: dx $ 9.2 The sketch below represents function $ f $ defined by $ f(x) = -x^2 + 6 $, with $ x $-intercepts at (-3; 0) and (2; 0) - NSC Technical Mathematics - Question 9 - 2020 - Paper 1

Question 9

$9.1-Determine-the-following:--9.1.1-$-\int-2x-\:-dx-$--9.1.2-$-\int-\left(-\sqrt{x}-+-\frac{7}{x}-+-4x^{-2}-\right)-\:-dx-$--9.2-The-sketch-below-represents-function-$-f-$-defined-by-$-f(x)-=--x^2-+-6-$,-with-$-x-$-intercepts-at-(-3;-0)-and-(2;-0)-NSC Technical Mathematics-Question 9-2020-Paper 1.png$

9.1 Determine the following: 9.1.1 $ \int 2x \: dx $ 9.1.2 $ \int \left( \sqrt{x} + \frac{7}{x} + 4x^{-2} \right) \: dx $ 9.2 The sketch below represents func... show full transcript

Worked Solution & Example Answer:9.1 Determine the following: 9.1.1 $ \int 2x \: dx $ 9.1.2 $ \int \left( \sqrt{x} + \frac{7}{x} + 4x^{-2} \right) \: dx $ 9.2 The sketch below represents function $ f $ defined by $ f(x) = -x^2 + 6 $, with $ x $-intercepts at (-3; 0) and (2; 0) - NSC Technical Mathematics - Question 9 - 2020 - Paper 1

Step 1

9.1.1 $ \int 2x \: dx $

96%

114 rated

Answer

To evaluate the integral ( \int 2x , dx ), we apply the power rule:

[ \int 2x , dx = 2 \cdot \frac{x^2}{2} + C = x^2 + C ]

Step 2

9.1.2 $ \int \left( \sqrt{x} + \frac{7}{x} + 4x^{-2} \right) \: dx $

99%

104 rated

Answer

This integral can be evaluated term by term:

[ \int \left( \sqrt{x} + \frac{7}{x} + 4x^{-2} \right) : dx = \int x^{1/2} , dx + \int 7x^{-1} , dx + \int 4x^{-2} , dx ]

Evaluating each integral:

( \int x^{1/2} , dx = \frac{x^{3/2}}{3/2} = \frac{2x^{3/2}}{3} )
( \int 7x^{-1} , dx = 7 \ln |x| )
( \int 4x^{-2} , dx = -\frac{4}{x} )

Combining these results:

[ \int \left( \sqrt{x} + \frac{7}{x} + 4x^{-2} \right) : dx = \frac{2x^{3/2}}{3} + 7 \ln |x| - \frac{4}{x} + C ]

Step 3

9.2 Determine whether the unshaded area, bounded by the curve and the x-axis between the points x = 3 and x = 2, is less than the shaded area.

96%

101 rated

Answer

First, compute the area above the x-axis from ( x = 2 ) to ( x = 3 ):

[ \int_{2}^{3} (-x^2 + 6) , dx = \left[ -\frac{x^3}{3} + 6x \right]_{2}^{3} ]

Calculating the definite integral:

[ = \left( -\frac{27}{3} + 18 \right) - \left( -\frac{8}{3} + 12 \right) = -9 + 18 + \frac{8}{3} - 12 = 6 - 9 + \frac{8}{3} = \frac{8}{3} \text{ square units} ]

The total unshaded area:

[ \text{Unshaded Area} = \frac{8}{3} \text{ square units} ]

Given the shaded area is ( \frac{34}{3} ) square units, therefore:

[ \frac{8}{3} < \frac{34}{3} \text{ hence, the unshaded area is less than the shaded area.} ]

9.1.1 \( \int 2x \: dx \)

9.1.2 \( \int \left( \sqrt{x} + \frac{7}{x} + 4x^{-2} \right) \: dx \)

9.2 Determine whether the unshaded area, bounded by the curve and the x-axis between the points x = 3 and x = 2, is less than the shaded area.

Join the NSC students using SimpleStudy...