Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2- x^3$. The curve has a maximum turning point $A$. (a) Using calculus, show that the x-coordinate of $A$ is 2. (b) The region $R$, shown shaded in Figure 2, is bounded by the curve, the y-axis and the line from $O$ to $A$, where $O$ is the origin. Using calculus, find the exact area of $R$.

A-Level Edexcel Maths Pure Integration: Figure 2 shows a sketch of part

Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2- x^3$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2

Question 8

Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2- x^3$. The curve has a maximum turning point $A$. (a) Using calculus, show that the x-coo... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2- x^3$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2

Step 1

Using calculus, show that the x-coordinate of A is 2.

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Answer

To find the x-coordinate of point $A$ , we first need to differentiate the equation of the curve:

$y = 10 + 8x + x^2 - x^3$

Differentiating with respect to $x$ , we have:

$\frac{dy}{dx} = 8 + 2x - 3x^2$

Setting this derivative to zero for finding critical points:

$8 + 2x - 3x^2 = 0$

Rearranging gives:

$3x^2 - 2x - 8 = 0$

Applying the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 3$ , $b = -2$ , and $c = -8$ :

$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 3 \times (-8)}}{2 \times 3} = \frac{2 \pm \sqrt{4 + 96}}{6} = \frac{2 \pm \sqrt{100}}{6}$

Thus:

$x = \frac{2 \pm 10}{6}$

Calculating the two possible values:

$x = \frac{12}{6} = 2$ (valid)
$x = \frac{-8}{6} = -\frac{4}{3}$ (not in the relevant region)

Therefore, the x-coordinate of turning point $A$ is 2.

Step 2

Using calculus, find the exact area of R.

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Answer

To find the area of the region $R$ , we can set up the integral of the curve from the origin ( $x = 0$ ) to point $A$ ( $x = 2$ ):

$\text{Area} = \int_0^2 (10 + 8x + x^2 - x^3) \, dx$

Calculating this integral:

Finding the antiderivative:
- The integral of $10$ is $10x$ .
- The integral of $8x$ is $4x^2$ .
- The integral of $x^2$ is $\frac{x^3}{3}$ .
- The integral of $-x^3$ is $-\frac{x^4}{4}$ .

Thus:

$\int (10 + 8x + x^2 - x^3) \, dx = 10x + 4x^2 + \frac{x^3}{3} - \frac{x^4}{4} + C$

Evaluating the definite integral:
- Now, we calculate:

$\text{Area} = \left[ 10x + 4x^2 + \frac{x^3}{3} - \frac{x^4}{4} \right]_0^2$

Substituting the limits:
- At $x = 2$ :
$\text{Area} = 10(2) + 4(2^2) + \frac{(2^3)}{3} - \frac{(2^4)}{4}$

$= 20 + 16 + \frac{8}{3} - 4$

$= 32 + \frac{8}{3}$

$= \frac{96}{3} + \frac{8}{3} = \frac{104}{3}$
Total area:
- Thus, the exact area of region $R$ is:
$\text{Area} = \frac{104}{3} \text{ units}^2$ .

Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2- x^3$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y=10+8x+x^2- x^3$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2

Using calculus, show that the x-coordinate of A is 2.

Using calculus, find the exact area of R.

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