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Figure 2 shows the line with equation $y = 10 - x$ and the curve with equation $y = 10x - x^2 - 8$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 3

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Figure-2-shows-the-line-with-equation-$y-=-10---x$-and-the-curve-with-equation-$y-=-10x---x^2---8$-Edexcel-A-Level Maths Pure-Question 6-2012-Paper 3.png

Figure 2 shows the line with equation $y = 10 - x$ and the curve with equation $y = 10x - x^2 - 8$. The line and the curve intersect at the points A and B, and O is... show full transcript

Worked Solution & Example Answer:Figure 2 shows the line with equation $y = 10 - x$ and the curve with equation $y = 10x - x^2 - 8$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 3

Step 1

Calculate the coordinates of A and the coordinates of B.

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Answer

To find the coordinates of points A and B, we need to solve the equations:

  1. Set the two equations equal to each other: 10−x=10x−x2−810 - x = 10x - x^2 - 8

  2. Rearrange this equation: x2−11x+18=0x^2 - 11x + 18 = 0

  3. Factor the quadratic: (x−2)(x−9)=0(x - 2)(x - 9) = 0

  4. The solutions give us: x=2extandx=9x = 2 ext{ and } x = 9

  5. To find the corresponding y-coordinates, substitute these x-values back into the line equation y=10−xy = 10 - x:

    • For x=2x = 2: y=10−2=8y = 10 - 2 = 8 Thus, point A is (2,8)(2, 8).
    • For x=9x = 9: y=10−9=1y = 10 - 9 = 1 Thus, point B is (9,1)(9, 1).

Step 2

Calculate the exact area of R.

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Answer

To calculate the area of region R, we use integration:

  1. Identify the top curve and the bottom curve:

    • Top curve: y=10x−x2−8y = 10x - x^2 - 8
    • Bottom curve: y=10−xy = 10 - x

  2. Determine the points of intersection (A and B), found as (2,8)(2, 8) and (9,1)(9, 1).

  3. Set up the integral for the area: extArea=∫29[(10x−x2−8)−(10−x)] dx ext{Area} = \int_{2}^{9} [(10x - x^2 - 8) - (10 - x)] \, dx

  4. Simplifying the integrand: (10x−x2−8−10+x)=−x2+11x−18(10x - x^2 - 8 - 10 + x) = -x^2 + 11x - 18

  5. Integrate: =∫29(−x2+11x−18) dx= \int_{2}^{9} (-x^2 + 11x - 18) \, dx =[−x33+11x22−18x]29= \left[-\frac{x^3}{3} + \frac{11x^2}{2} - 18x\right]_{2}^{9}

  6. Evaluate the definite integral:

    • Calculate at x=9x = 9: =−933+11(92)2−18(9)= -\frac{9^3}{3} + \frac{11(9^2)}{2} - 18(9) =−243+445.5−162=40.5= -243 + 445.5 - 162 = 40.5

    • Calculate at x=2x = 2: =−233+11(22)2−18(2)= -\frac{2^3}{3} + \frac{11(2^2)}{2} - 18(2) =−83+44−36=8−83=24−83=163= -\frac{8}{3} + 44 - 36 = 8 - \frac{8}{3} = \frac{24 - 8}{3} = \frac{16}{3}

  7. Final area calculation: Area=40.5−163=121.5−163=105.53=35.1667\text{Area} = 40.5 - \frac{16}{3} = \frac{121.5 - 16}{3} = \frac{105.5}{3} = 35.1667 .

The exact area of R is 105.53\frac{105.5}{3} square units.

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