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Figure 1 shows part of the curve with equation $y = e^{0.5x}$ - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 7

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Figure 1 shows part of the curve with equation $y = e^{0.5x}$. The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the x-axis, the y-axis and t... show full transcript

Worked Solution & Example Answer:Figure 1 shows part of the curve with equation $y = e^{0.5x}$ - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 7

Step 1

Complete the table with the values of $y$ corresponding to $x = 0.8$ and $x = 1.6$

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Answer

xxyy
011
0.41.49181.4918
0.82.22552.2255
1.23.32013.3201
1.64.95304.9530
27.38917.3891

Step 2

Use the trapezium rule to find an approximate value for the area of $R$

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Answer

To apply the trapezium rule:

  1. Identify the values of yy: y0=1y_0 = 1, y1=1.4918y_1 = 1.4918, y2=2.2255y_2 = 2.2255, y3=3.3201y_3 = 3.3201, y4=4.9530y_4 = 4.9530, y5=7.3891y_5 = 7.3891.

  2. The area is calculated as:

    ext{Area} = rac{1}{2} imes ext{interval width} imes (y_0 + 2y_1 + 2y_2 + 2y_3 + 2y_4 + y_5)

  3. The interval width is 0.40.4;

    ext{Area} = rac{1}{2} imes 0.4 imes (1 + 2(1.4918) + 2(2.2255) + 2(3.3201) + 2(4.9530) + 7.3891)

  4. Performing the calculations:

    extArea=0.2imes(1+2.9836+4.4510+6.6402+9.9060+7.3891)=0.2imes32.3709=4.4744 ext{Area} = 0.2 imes (1 + 2.9836 + 4.4510 + 6.6402 + 9.9060 + 7.3891) = 0.2 imes 32.3709 = 4.4744

  5. Rounding to 4 significant figures gives:

    extAreaextofRextisapproximately4.474 ext{Area} ext{ of } R ext{ is approximately } 4.474.

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