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Complete the table with the values of y corresponding to x = 2 and 2.5 - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 4

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Complete the table with the values of y corresponding to x = 2 and 2.5. | x | y | | ---- | ---- | | 1.0 | 16.5 | | 1.5 | 7.361| | 2 | 4 | | 2.5 | 2.31... show full transcript

Worked Solution & Example Answer:Complete the table with the values of y corresponding to x = 2 and 2.5 - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 4

Step 1

Complete the table with the values of y corresponding to x = 2 and 2.5.

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Answer

For x = 2, substituting into the equation:

y=162222+1=41+1=4y = \frac{16}{2^2} - \frac{2}{2} + 1 = 4 - 1 + 1 = 4

For x = 2.5, substituting into the equation:

y=16(2.5)22.52+1=166.251.25+1=2.561.25+1=2.31y = \frac{16}{(2.5)^2} - \frac{2.5}{2} + 1 = \frac{16}{6.25} - 1.25 + 1 = 2.56 - 1.25 + 1 = 2.31

Step 2

Use the trapezium rule with all the values in the completed table to find an approximate value for the area of R, giving your answer to 2 decimal places.

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Answer

Using the trapezium rule, the formula for the area is:

A=12×(b1+bn)+i=1n1biA = \frac{1}{2} \times (b_1 + b_n) + \sum_{i=1}^{n-1} b_i

Where b are the y-values. Here b_1 = 16.5, b_n = 0, and the other y-values are: 7.361, 4, 2.31, 1.278, 0.556.

Calculating:

A=12(16.5+0)+(7.361+4+2.31+1.278+0.556)A = \frac{1}{2}(16.5 + 0) + (7.361 + 4 + 2.31 + 1.278 + 0.556) A=8.25+31.505=39.755A = 8.25 + 31.505 = 39.755

Thus, the approximate area is 11.88 (to 2 decimal places).

Step 3

Use integration to find the exact value for the area of R.

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Answer

To find the exact area of R, we integrate the function:

A=14(16x2x2+1) dxA = \int_{1}^{4} \left( \frac{16}{x^2} - \frac{x}{2} + 1 \right) \ dx

Calculating the integral:

=[16xx24+x]14= \left[ -\frac{16}{x} - \frac{x^2}{4} + x \right]_{1}^{4}

Evaluating the bounds:

=[44+4][1614+1]= \left[ -4 - 4 + 4 \right] - \left[ -16 - \frac{1}{4} + 1 \right] =[4][16.25]=12.25= [ -4 ] - [ -16.25 ] = 12.25

Thus, the exact area is 12.25.

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