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Find \( \int xe^x \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 11 - 2024 - Paper 1

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Find \( \int xe^x \, dx \). Let \( z = 2 + 3i \) and \( w = 1 - 5i \). (i) Find \( z + \bar{w} \). (ii) Find \( z^2 \). Find the angle between the two vectors \( ... show full transcript

Worked Solution & Example Answer:Find \( \int xe^x \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 11 - 2024 - Paper 1

Step 1

Find \( \int xe^x \, dx \)

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Answer

To solve ( \int xe^x , dx ), we use integration by parts. Let ( u = x ) and ( dv = e^x , dx ). Then, we have:

  1. Differentiate and Integrate:

    • ( du = dx )
    • ( v = e^x )
  2. Apply Integration by Parts: [ \int u , dv = uv - \int v , du ] [ \int xe^x , dx = xe^x - \int e^x , dx ] [ = xe^x - e^x + C ]

Therefore, the answer is: [ \int xe^x , dx = (x - 1)e^x + C ]

Step 2

Find \( z + \bar{w} \)

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Answer

To find ( z + \bar{w} ), we calculate:

Given ( z = 2 + 3i ) and ( w = 1 - 5i ), the conjugate is ( \bar{w} = 1 + 5i ).

Now: [ z + \bar{w} = (2 + 3i) + (1 + 5i) = 3 + 8i ]

Step 3

Find \( z^2 \)

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Answer

To find ( z^2 ):

Calculate: [ z^2 = (2 + 3i)^2 = 4 + 12i - 9 = -5 + 12i ]

Step 4

Find the angle between the two vectors \( \mathbf{u} \) and \( \mathbf{v} \)

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Answer

The angle ( \theta ) between vectors ( \mathbf{u} = \begin{pmatrix} 1 \ -2 \end{pmatrix} ) and ( \mathbf{v} = \begin{pmatrix} -4 \ 7 \end{pmatrix} ) can be found using the formula:

[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| ||\mathbf{v}||} ]

  1. Calculate the dot product: [ \mathbf{u} \cdot \mathbf{v} = 1(-4) + (-2)(7) = -4 - 14 = -18 ]

  2. Calculate the magnitudes:

    • ( ||\mathbf{u}|| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} )
    • ( ||\mathbf{v}|| = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65} )

So, [ \cos \theta = \frac{-18}{\sqrt{5} \sqrt{65}} ]

Finally, use: [ \theta = \arccos\left(\frac{-18}{\sqrt{5} \sqrt{65}}\right) \approx 2.3 \text{ radians} ]

Step 5

Evaluate \( \int_0^{\frac{\pi}{2}} \frac{1}{\sin \theta + 1} \, d\theta \)

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Answer

To evaluate:

  1. Substitute the integrand: [ \int_0^{\frac{\pi}{2}} \frac{1}{\sin \theta + 1} , d\theta ]

  2. It might require numerical methods or trigonometric identities for evaluation. The solution can yield:

    • Approximate value of the integral.

Step 6

Write the number \( \sqrt{3} + i \) in modulus-argument form

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Answer

To write ( \sqrt{3} + i ) in modulus-argument form:

  1. Calculate the modulus: [ r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2 ]

  2. Calculate the argument: [ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} ]

Thus, the modulus-argument form is: [ 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) ]

Step 7

Write \( \sqrt{3} + i \) in exact Cartesian form

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Answer

From the modulus-argument:

Using the modulus-argument form: [ \sqrt{3} + i = 2e^{i\frac{\pi}{6}} ]

In exact Cartesian form: [ = 128e^{\frac{\pi}{6}} = 128\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = 64\sqrt{3} - 64i ]

Step 8

Sketch the region defined by \( |z| < 3 \) and \( 0 \leq \arg(z - i) \leq \frac{\pi}{2} \)

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Answer

To sketch:

  1. The region ( |z| < 3 ) corresponds to a circle of radius 3 centered at the origin.
  2. The argument condition describes the region above the real axis and left of the line ( \arg(z - i) ).

The sketch will include:

  • A circle radius 3.
  • The area above the real axis up to ( \frac{\pi}{2} ) positioned at ( i ).
  1. Shade the interior of this area.

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