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Figure 1 shows the finite region R bounded by the x-axis, the y-axis, the line x = \frac{\pi}{2} and the curve with equation y = sec(\frac{1}{2} x), \quad 0 \leq x \leq \frac{\pi}{2} - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 9
Question 5
Figure 1 shows the finite region R bounded by the x-axis, the y-axis, the line x = \frac{\pi}{2} and the curve with equation y = sec(\frac{1}{2} x), \quad 0 \leq x ... show full transcript
Worked Solution & Example Answer:Figure 1 shows the finite region R bounded by the x-axis, the y-axis, the line x = \frac{\pi}{2} and the curve with equation y = sec(\frac{1}{2} x), \quad 0 \leq x \leq \frac{\pi}{2} - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 9
Step 1
Complete the table above giving the missing value of y to 6 decimal places.
Answer
To complete the table, we need to calculate the value of y when x = \frac{\pi}{3}:
Using the given formula:
y = sec(\frac{1}{2} x) = sec(\frac{1}{2} \cdot \frac{\pi}{3}) = sec(\frac{\pi}{6})
Secant can be computed as:
sec(\frac{\pi}{6}) = \frac{1}{\cos(\frac{\pi}{6})} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \approx 1.154701
Thus, the completed table is:
| x | 0 | \frac{\pi}{6} | \frac{\pi}{3} | \frac{\pi}{2} |
|---|---|---|---|---|
| y | 1 | 1.035276 | 1.154701 | 1.414214 |
Step 2
Using the trapezium rule, with all of the values of y from the completed table, find an approximation for the area of R, giving your answer to 4 decimal places.
Answer
To approximate the area of region R using the trapezium rule, we apply the formula:
[ A = \frac{h}{2} \left( f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3) \right) ]
Here, ( h = \frac{\pi/2 - 0}{3} = \frac{\pi}{6} ) and the function values are:
- f(0) = 1
- f(\frac{\pi}{6}) = 1.035276
- f(\frac{\pi}{3}) = 1.154701
- f(\frac{\pi}{2}) = 1.414214
Substituting the values:
[ A = \frac{\frac{\pi}{6}}{2} \left( 1 + 2(1.035276) + 2(1.154701) + 1.414214 \right) ]
Performing the calculations,
[ A = \frac{\frac{\pi}{6}}{2} \left( 1 + 2.070552 + 2.309402 + 1.414214 \right) ]
[ A = \frac{\frac{\pi}{6}}{2} \left( 6.794168 \right) ]
Finally, computing the area:
[ A \approx \frac{\pi}{12} \cdot 6.794168 \approx 0.597012 \cdot \pi \approx 1.877075392 \text{ (to four decimal places: } 1.7787 \text{)} ]
Step 3
Region R is rotated around 2\pi radians about the x-axis.
Answer
To find the volume of the solid formed when region R is rotated about the x-axis, we apply the disk method:
[ V = \pi \int_{0}^{\frac{\pi}{2}} (y^2) , dx ]
Given ( y = sec(\frac{1}{2} x) ), we compute:
[ V = \pi \int_{0}^{\frac{\pi}{2}} (sec(\frac{1}{2} x))^2 , dx ]
Using the identity ( sec^2(u) = 1 + \tan^2(u) ):
[ V = \pi \left[ 2tan(\frac{1}{2} x) \right]_{0}^{\frac{\pi}{2}} ]
Calculating the limits:
- At ( x = \frac{\pi}{2} ): ( tan(\frac{\pi}{4}) = 1 )
- At ( x = 0 ): ( tan(0) = 0 )
Thus, the volume is:
[ V = \pi [2(1) - 0] = 2\pi ]
Hence, the volume of the solid formed is ( 2\pi ).