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Find \( \int xe^x \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 11 - 2024 - Paper 1
Question 11
Find \( \int xe^x \, dx \). Let \( z = 2 + 3i \) and \( w = 1 - 5i \). (i) Find \( z + \overline{w} \). (ii) Find \( z^2 \). Find the angle between the two vecto... 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 \( \int xe^x \, dx \)
Answer
To find the integral ( \int xe^x , dx ), we use integration by parts. Let:
- ( u = x )
- ( dv = e^x , dx )
Then, differentiate and integrate:
- ( du = dx )
- ( v = e^x )
Applying the integration by parts formula ( \int u , dv = uv - \int v , du ):
[ \int xe^x , dx = xe^x - \int e^x , dx = xe^x - e^x + C. ]
Thus, the answer is ( xe^x - e^x + C ).
Step 2
Step 3
Step 4
Find the angle between the two vectors \( \mathbf{u} \) and \( \mathbf{v} \)
Answer
Using the formula for the angle between two vectors:
[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} ]
First, calculate the dot product:
[ \mathbf{u} \cdot \mathbf{v} = 1 \cdot (-4) + (-2) \cdot 7 = -4 - 14 = -18. ]
Now calculate the magnitudes:
[ |\mathbf{u}| = \sqrt{1^2 + (-2)^2} = \sqrt{5}, \quad |\mathbf{v}| = \sqrt{(-4)^2 + 7^2} = \sqrt{65}. ]
Substituting into the formula gives:
[ \cos \theta = \frac{-18}{\sqrt{5} \sqrt{65}} \approx -0.721. ]
Thus, ( \theta ) is:
[ \theta \approx \cos^{-1}(-0.721) \approx 2.3 \text{ radians}. ]
Step 5
Evaluate \( \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin \theta + 1} \, d\theta. \)
Answer
We make the substitution ( t = \sin \theta + 1 ), then ( dt = \cos \theta , d\theta ). Changing the limits from ( \theta = 0 ) to ( \theta = \frac{\pi}{2} ):
- When ( \theta = 0 ), ( t = 1 )
- When ( \theta = \frac{\pi}{2} ), ( t = 2 )
This transforms our integral:
[ \int_{1}^{2} \frac{1}{t} , dt = \ln t \Big|_{1}^{2} = \ln 2. ]
Step 6
Write the number \( \sqrt{3} + i \) in modulus-argument form.
Answer
To express ( \sqrt{3} + i ) in modulus-argument form, we first find the modulus:
[ r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2. ]
Next, we find the argument:
[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}. ]
Thus, the modulus-argument form is:
[ \sqrt{3} + i = 2 \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right). ]
Step 7
Step 8
Sketch the region defined by \( |z| < 3 \) and \( 0 \leq \arg(z - i) \leq \frac{\pi}{2}. \)
Answer
The region defined by ( |z| < 3 ) is the interior of a circle with radius 3 centered at the origin. The condition ( 0 \leq \arg(z - i) \leq \frac{\pi}{2} ) restricts the region to the first quadrant. Therefore, the sketch shows a circular sector with radius 3 in the first quadrant, bounded by the positive real axis (where ( \arg = 0 )) and the line ( y = x ) (where ( \arg = \frac{\pi}{2} )).