The line with equation $y = 10$ cuts the curve with equation $y = x^2 + 2x + 2$ at the points A and B as shown in Figure 1. The figure is not drawn to scale. (a) Find by calculation the x-coordinate of A and the x-coordinate of B. (b) The shaded region R is bounded by the line with equation $y = 10$ and the curve as shown in Figure 1.

A-Level Edexcel Maths Pure Momentum: The line with equation $y = 10$

The line with equation $y = 10$ cuts the curve with equation $y = x^2 + 2x + 2$ at the points A and B as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 5

Question 8

The line with equation $y = 10$ cuts the curve with equation $y = x^2 + 2x + 2$ at the points A and B as shown in Figure 1. The figure is not drawn to scale. (a) Fi... show full transcript

Worked Solution & Example Answer:The line with equation $y = 10$ cuts the curve with equation $y = x^2 + 2x + 2$ at the points A and B as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 5

Step 1

Find by calculation the x-coordinate of A and the x-coordinate of B.

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Answer

To find the x-coordinates of points A and B, we set the equations equal to each other:

$x^2 + 2x + 2 = 10$

This simplifies to:

$x^2 + 2x - 8 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$ , $b = 2$ , and $c = -8$ ,

we find:

$b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36$

Thus, the roots are:

$x = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}$

Calculating gives:

$x_1 = 2$ $x_2 = -4$

Therefore, the x-coordinates of A and B are $x = 2$ and $x = -4$ , respectively.

Step 2

Use calculus to find the exact area of R.

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Answer

To find the area of the shaded region R, we calculate the integral between the x-coordinates found: $-4$ and $2$ .

The area A can be calculated as follows:

$A = \int_{-4}^{2} (10 - (x^2 + 2x + 2)) \, dx$

This simplifies to:

$A = \int_{-4}^{2} (10 - x^2 - 2x - 2) \, dx$

Which further simplifies to:

$A = \int_{-4}^{2} (8 - x^2 - 2x) \, dx$

Calculating the integral:

$\int (8 - x^2 - 2x) \, dx = 8x - \frac{x^3}{3} - x^2 + C$

Evaluate this from $-4$ to $2$ :

At $x = 2$ : $8(2) - \frac{(2)^3}{3} - (2)^2 = 16 - \frac{8}{3} - 4 = 12 - \frac{8}{3} = \frac{36 - 8}{3} = \frac{28}{3}$
At $x = -4$ : $8(-4) - \frac{(-4)^3}{3} - (-4)^2 = -32 + \frac{64}{3} - 16 = -48 + \frac{64}{3} = \frac{-144 + 64}{3} = \frac{-80}{3}$

Now subtract the two results:

$A = \left(\frac{28}{3} - \frac{-80}{3}\right) = \frac{28}{3} + \frac{80}{3} = \frac{108}{3} = 36$

Thus, the exact area of region R is $36$ .

The line with equation $y = 10$ cuts the curve with equation $y = x^2 + 2x + 2$ at the points A and B as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 5

Worked Solution & Example Answer:The line with equation $y = 10$ cuts the curve with equation $y = x^2 + 2x + 2$ at the points A and B as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 5

Find by calculation the x-coordinate of A and the x-coordinate of B.

Use calculus to find the exact area of R.

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