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 the origin. (a) Calculate the coordinates of A and the coordinates of B. (b) The shaded area R is bounded by the line and the curve, as shown in Figure 2. Calculate the exact area of R.

A-Level Edexcel Maths Pure Trigonometric Equations: Figure 2 shows the line with equ

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

Question 6

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 where the line and curve intersect, we set the equations equal to each other:

$10 - x = 10x - x^2 - 8$

Rearranging the equation gives:

$x^2 - 11x + 18 = 0$

Next, we can factor this quadratic equation as:

$(x - 2)(x - 9) = 0$

This gives us the solutions:

$x = 2\text{ and } x = 9$

Now, substituting these $x$ values back into the line equation to find the respective $y$ coordinates:

For $x = 2$ :
$y = 10 - 2 = 8$
So point A is $(2, 8)$ .
For $x = 9$ : $y = 10 - 9 = 1$ So point B is $(9, 1)$ .

Thus, the coordinates of A are (2, 8) and the coordinates of B are (9, 1).

Step 2

Calculate the exact area of R.

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Answer

To calculate the area of the shaded region R, we need to find the area between the line and the curve from x = 2 to x = 9. The area A can be calculated using the integral:

$A = \int_{2}^{9} ((10 - x) - (10x - x^2 - 8)) \, dx$

This simplifies to:

$A = \int_{2}^{9} (x^2 - 9x + 18) \, dx$

Now, we can calculate the integral:

$A = \left[ \frac{x^3}{3} - \frac{9x^2}{2} + 18x \right]_{2}^{9}$

Calculating the definite integral:

At $x = 9$ : $\frac{9^3}{3} - \frac{9(9^2)}{2} + 18(9) = 243 - 364.5 + 162 = 40.5$

At $x = 2$ : $\frac{2^3}{3} - \frac{9(2^2)}{2} + 18(2) = \frac{8}{3} - 18 + 36 = \frac{8}{3} + 18 = \frac{62}{3}$

Thus:

$A = 40.5 - \frac{62}{3} = \frac{121.5 - 62}{3} = \frac{59.5}{3} = 19.8333$

The exact area of region R is approximately 19.83 square units.

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

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

Calculate the coordinates of A and the coordinates of B.

Calculate the exact area of R.

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