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 Initial Conditions: 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 where the line $y = 10$ intersects the curve $y = x^2 + 2x + 2$ , we set these equations equal:

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

Rearranging gives:

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

Next, we can use the quadratic formula where $a = 1$ , $b = 2$ , and $c = -8$ :

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)}$

This simplifies to: $x = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}$

From this we have:

$x = \frac{4}{2} = 2$ $x = \frac{-8}{2} = -4$

Thus, the x-coordinates of points 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 calculate the area of the region R, we first set up the integral of the function $y = x^2 + 2x + 2$ from $x = -4$ to $x = 2$ and subtracting it from the rectangle formed under the line $y = 10$ :

The area of the rectangle is: $\text{Area of rectangle} = \text{Base} \times \text{Height} = (2 - (-4)) \times 10 = 6 \times 10 = 60$

Now, the area under the curve from $x = -4$ to $x = 2$ is calculated as follows:

$\int_{-4}^{2} (10 - (x^2 + 2x + 2)) \, dx = \int_{-4}^{2} (8 - x^2 - 2x) \, dx$

Evaluating the integral, we have:

$[8x - \frac{x^3}{3} - x^2]_{-4}^{2}$

Calculating at the bounds: At $x = 2$ : $8(2) - \frac{(2)^3}{3} - (2)^2 = 16 - \frac{8}{3} - 4 = 12 - \frac{8}{3} = \frac{36}{3} - \frac{8}{3} = \frac{28}{3}$

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 we subtract:

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

Finally, the area R is: $\text{Area of R} = \text{Area of rectangle} - \text{Area under curve} = 60 - 36 = 24$

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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