A-Level Edexcel Maths Pure Trigonometric Equations: 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 9 - 2013 - Paper 5

Question 9

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... 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 9 - 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 $B$ , set the equations equal to each other:

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

Rearranging gives:

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

Using the quadratic formula:

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

Here, $a = 1$ , $b = 2$ , and $c = -8$ :

Calculate the discriminant: $b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36$
Applying the quadratic formula: $x = \frac{-2 \pm 6}{2}$

This gives two solutions:

For $A$ : $x = \frac{4}{2} = 2$
For $B$ : $x = \frac{-8}{2} = -4$

Step 2

Use calculus to find the exact area of R.

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Answer

To find the area of the shaded region $R$ :

Identify the integral bounds from $x = -4$ to $x = 2$ .
Calculate the area using the integral:

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

This simplifies to:

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

Now compute the integral:

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

Evaluating at the bounds:
- At $x = 2$ : $8(2) - \frac{2^3}{3} - (2)^2 = 16 - \frac{8}{3} - 4 = 12 - \frac{8}{3} = \frac{36 - 8}{3} = \frac{28}{3}$
- At $x = -4$ : $8(-4) - \frac{(-4)^3}{3} - (-4)^2 = -32 + \frac{64}{3} - 16 = -48 + \frac{64}{3} = -\frac{144}{3} + \frac{64}{3} = -\frac{80}{3}$
Now, subtract:

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

Thus, the area of the 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 9 - 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 9 - 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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