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8. (a) Find \( \int \limits_{0}^{\frac{\pi}{4}} x \cos 4x \, dx \) Figure 3 shows part of the curve with equation \( y = \sqrt{x} \sin 2x, \; x > 0 \) - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 9
Question 1
8. (a) Find \( \int \limits_{0}^{\frac{\pi}{4}} x \cos 4x \, dx \) Figure 3 shows part of the curve with equation \( y = \sqrt{x} \sin 2x, \; x > 0 \). The finite ... show full transcript
Worked Solution & Example Answer:8. (a) Find \( \int \limits_{0}^{\frac{\pi}{4}} x \cos 4x \, dx \) Figure 3 shows part of the curve with equation \( y = \sqrt{x} \sin 2x, \; x > 0 \) - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 9
Step 1
Find \( \int \limits_{0}^{\frac{\pi}{4}} x \cos 4x \, dx \)
Answer
To find this integral, we can use integration by parts. Let:
- ( u = x ) and ( dv = \cos 4x , dx )
Thus,[ du = dx
v = \frac{1}{4} \sin 4x ]
Using the integration by parts formula ( \int u , dv = uv - \int v , du ), we have: [ \int x \cos 4x , dx = \left( x \cdot \frac{1}{4} \sin 4x \right) - \int \frac{1}{4} \sin 4x , dx ]
Now, compute ( \int \sin 4x , dx ): [ \int \sin 4x , dx = -\frac{1}{4} \cos 4x + C ]
Substituting this back: [ \int x \cos 4x , dx = \left( x \cdot \frac{1}{4} \sin 4x \right) + \frac{1}{16} \cos 4x + C ]
Evaluating from ( 0 ) to ( \frac{\pi}{4} ): [ = \left[ \frac{\pi}{16} \sin \left( 4 \cdot \frac{\pi}{4} \right) + \frac{1}{16} \cos \left( 4 \cdot \frac{\pi}{4} \right) \right] - \left[ 0 + \frac{1}{16} \cos(0) \right] ] [ = \left[ \frac{\pi}{16} \sin \pi + \frac{1}{16} \cos \pi \right] - \frac{1}{16} = 0 + \left( -\frac{1}{16} \right) - \frac{1}{16} ] [ = -\frac{2}{16} = -\frac{1}{8} ]
Step 2
Find the exact value of the volume of this solid of revolution
Answer
To find the volume of the solid of revolution formed by rotating region ( R ) about the x-axis, we use the formula: [ V = \pi \int_{a}^{b} [f(x)]^2 , dx ] where ( f(x) = \sqrt{x} \sin 2x ) and in our case: Thus: [ V = \pi \int_0^{\frac{\pi}{4}} (\sqrt{x} \sin 2x)^2 , dx = \pi \int_0^{\frac{\pi}{4}} x \sin^2 2x , dx ]
Using the identity ( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} ), we rewrite the integral: [ = \frac{\pi}{2} \left( \int_0^{\frac{\pi}{4}} x , dx - \int_0^{\frac{\pi}{4}} x \cos 4x , dx \right) ]
Now, compute the integral ( \int_0^{\frac{\pi}{4}} x , dx ): [ = \frac{x^2}{2} \bigg|_0^{\frac{\pi}{4}} = \frac{\left( \frac{\pi}{4} \right)^2}{2} = \frac{\pi^2}{32} ]
Finally, we substitute this back into the volume integral and evaluate: [ V = \frac{\pi}{2} \left( \frac{\pi^2}{32} - \left(-\frac{1}{8} \right) \right) ] Calculating gives the final volume value as: [ V = \frac{\pi^3}{64} + \frac{\pi}{16} = \frac{\pi(\pi^2 + 4)}{64} ]