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Bepaal die volgende integral: 9.1.1 ∫ 3/x dx 9.1.2 ∫ (3x/x^5 + √x^3) dx 9.2 Die skets hieronder toon funksie f gedefinieer deur f(x)=x²−5x Die twee geaserde oppervlaktes wat voorgestel word, is: • A₁ = oppervlakte begrens deur funksie f, die x-as en die ordinates x=0 en x=3 • A₂ = oppervlakte begrens deur funksie f, die x-as en die ordinates x=5 en x=7 Bepaal (toon ALLE berekeninge) met hoeveel A₁ groter as A₂ is. - NSC Technical Mathematics - Question 9 - 2023 - Paper 1

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Question 9

Bepaal-die-volgende-integral:--9.1.1--∫-3/x--dx----9.1.2--∫-(3x/x^5-+-√x^3)-dx----9.2--Die-skets-hieronder-toon-funksie--f--gedefinieer-deur--f(x)=x²−5x----Die-twee-geaserde-oppervlaktes-wat-voorgestel-word,-is:-•--A₁-=-oppervlakte-begrens-deur-funksie--f,--die-x-as-en-die-ordinates--x=0--en--x=3---•--A₂-=-oppervlakte-begrens-deur-funksie--f,--die-x-as-en-die-ordinates--x=5--en--x=7----Bepaal-(toon-ALLE-berekeninge)-met-hoeveel--A₁--groter-as--A₂--is.-NSC Technical Mathematics-Question 9-2023-Paper 1.png

Bepaal die volgende integral: 9.1.1 ∫ 3/x dx 9.1.2 ∫ (3x/x^5 + √x^3) dx 9.2 Die skets hieronder toon funksie f gedefinieer deur f(x)=x²−5x Die twee ... show full transcript

Worked Solution & Example Answer:Bepaal die volgende integral: 9.1.1 ∫ 3/x dx 9.1.2 ∫ (3x/x^5 + √x^3) dx 9.2 Die skets hieronder toon funksie f gedefinieer deur f(x)=x²−5x Die twee geaserde oppervlaktes wat voorgestel word, is: • A₁ = oppervlakte begrens deur funksie f, die x-as en die ordinates x=0 en x=3 • A₂ = oppervlakte begrens deur funksie f, die x-as en die ordinates x=5 en x=7 Bepaal (toon ALLE berekeninge) met hoeveel A₁ groter as A₂ is. - NSC Technical Mathematics - Question 9 - 2023 - Paper 1

Step 1

9.1.1 ∫ 3/x dx

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Answer

To solve the integral, we recognize that it can be expressed as:

∫3/x dx=3∫1/x dx=3ln⁡∣x∣+C∫ 3/x \, dx = 3 ∫ 1/x \, dx = 3 \ln|x| + C

This gives us the answer: 3ln|x| + C.

Step 2

9.1.2 ∫ (3x/x^5 + √x^3) dx

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Answer

First, simplify the integrand:

rac{3x}{x^5} + \sqrt{x^3} = 3 x^{-4} + x^{3/2}

Then, we integrate each term separately:

∫(3x−4)dx=−33x−3=−x−3∫ (3 x^{-4}) dx = -\frac{3}{3} x^{-3} = -x^{-3}

and

∫x3/2 dx=25x5/2∫ x^{3/2} \, dx = \frac{2}{5} x^{5/2}

Thus,

∫(3x/x5+√x3)dx=−x−3+25x5/2+C∫ (3x/x^5 + √x^3) dx = -x^{-3} + \frac{2}{5} x^{5/2} + C

Step 3

A₁ = ∫ (x² - 5x) dx from x=0 to x=3

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Answer

To find A₁, we calculate:

A1=−∫(x2−5x)dx=−[x33−5x22]03A₁ = -∫ (x^2 - 5x) dx = -\left[ \frac{x^3}{3} - \frac{5x^2}{2} \right]_{0}^{3}

Calculating at the bounds:

=−[(3)33−5(3)22]−0=−[9−22.5]=13.5 units2= -\left[ \frac{(3)^3}{3} - \frac{5(3)^2}{2} \right] - 0 = -\left[ 9 - 22.5 \right] = 13.5 \, \text{units}^2

Step 4

A₂ = ∫ (x² - 5x) dx from x=5 to x=7

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Answer

Now, we calculate A₂:

A2=−∫(x2−5x)dx=−[x33−5x22]57A₂ = -∫ (x^2 - 5x) dx = -\left[ \frac{x^3}{3} - \frac{5x^2}{2} \right]_{5}^{7}

Plugging in the bounds:

=−[((7)33−5(7)22)−((5)33−5(5)22)]= -\left[ \left(\frac{(7)^3}{3} - \frac{5(7)^2}{2}\right) - \left(\frac{(5)^3}{3} - \frac{5(5)^2}{2}\right) \right]

Calculating,

=−[3433−5(49)2−1253+5(25)2]= -\left[ \frac{343}{3} - \frac{5(49)}{2} - \frac{125}{3} + \frac{5(25)}{2} \right]

This results in A₂ being 12.67 units².

Step 5

Vergelyk A₁ en A₂

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Answer

Finally, to find how much larger A₁ is than A₂:

A1−A2=13.5−12.67=0.83 units2A₁ - A₂ = 13.5 - 12.67 = 0.83 \, \text{units}^2

Thus, A₁ is greater than A₂ by 0.83 units².

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