Figure 3 shows a sketch of part of the curve with equation $y = x^3 - 8x^2 + 20x$. The curve has stationary points A and B. (a) Use calculus to find the x-coordinates of A and B. (b) Find the value of $ \frac{d^2y}{dx^2} $ at A, and hence verify that A is a maximum. (c) The line through B parallel to the y-axis meets the x-axis at the point N. The region R, shown shaded in Figure 3, is bounded by the curve, the x-axis and the line from A to N. (d) Find $ \int (x^3 - 8x^2 + 20x) \, dx $. (e) Hence calculate the exact area of R.

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Figure 3 shows a sketch of part of the curve with equation $y = x^3 - 8x^2 + 20x$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 2

Question 2

Figure 3 shows a sketch of part of the curve with equation $y = x^3 - 8x^2 + 20x$. The curve has stationary points A and B. (a) Use calculus to find the x-coordinat... show full transcript

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation $y = x^3 - 8x^2 + 20x$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 2

Step 1

Use calculus to find the x-coordinates of A and B.

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Answer

To find the stationary points A and B of the function, we start with the first derivative:

$\frac{dy}{dx} = 3x^2 - 16x + 20.$
Setting this equal to zero gives the equation:

$3x^2 - 16x + 20 = 0.$
Using the quadratic formula, where ( a = 3, b = -16, c = 20 ):

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot 20}}{2 \cdot 3} = \frac{16 \pm \sqrt{64}}{6} = \frac{16 \pm 8}{6}.$
Calculating the two potential solutions:

( x = \frac{24}{6} = 4 )
( x = \frac{8}{6} = \frac{4}{3} )

Thus, the x-coordinates of A and B are 4 and ( \frac{4}{3} ) respectively.

Step 2

Find the value of $ \frac{d^2y}{dx^2} $ at A, and hence verify that A is a maximum.

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Answer

To find the second derivative, we start with:

$\frac{dy}{dx} = 3x^2 - 16x + 20.$
Then the second derivative is:

$\frac{d^2y}{dx^2} = 6x - 16.$
Now substituting ( x = 4 ):

$\frac{d^2y}{dx^2}(4) = 6(4) - 16 = 24 - 16 = 8.$
Since ( \frac{d^2y}{dx^2} > 0 ), it confirms that A is a local maximum.

Step 3

Find $ \int (x^3 - 8x^2 + 20x) \, dx $.

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Answer

Integrating the function:

$\int (x^3 - 8x^2 + 20x) \, dx = \frac{x^4}{4} - \frac{8x^3}{3} + 10x^2 + C.$

where C is the constant of integration.

Step 4

Hence calculate the exact area of R.

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Answer

To find the exact area of the region R, we evaluate the definite integral from A to B:

The limits for the integral are from ( x = \frac{4}{3} ) to ( x = 4 ).
The area A is calculated as:

$Area = \int_{\frac{4}{3}}^{4} (x^3 - 8x^2 + 20x) \, dx = \left[ \frac{x^4}{4} - \frac{8x^3}{3} + 10x^2 \right]_{\frac{4}{3}}^{4}.$
Calculating the definite integral gives:

$A = \left(\frac{4^4}{4} - \frac{8 \cdot 4^3}{3} + 10 \cdot 16\right) - \left( \frac{(\frac{4}{3})^4}{4} - \frac{8 \cdot (\frac{4}{3})^3}{3} + 10 \cdot (\frac{4}{3})^2 \right).$

After evaluating, the final area of region R is obtained as ( \frac{32}{3} ).

Figure 3 shows a sketch of part of the curve with equation $y = x^3 - 8x^2 + 20x$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 2

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation $y = x^3 - 8x^2 + 20x$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 2

Use calculus to find the x-coordinates of A and B.

Find the value of \( \frac{d^2y}{dx^2} \) at A, and hence verify that A is a maximum.

Find \( \int (x^3 - 8x^2 + 20x) \, dx \).

Hence calculate the exact area of R.

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