Figure 2 shows a sketch of part of the curve with equation y = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A. (a) Using calculus, show that the x coordinate of A is 1. The curve crosses the x-axis at the points B(2, 0) and C(-\frac{1}{4}, 0). The finite region R, shown shaded in Figure 2, is bounded by the curve, the line AB, and the x-axis. (b) Use integration to find the finite region R, giving your answer to 2 decimal places.

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 = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 1 - 2016 - Paper 3

Question 1

Figure-2-shows-a-sketch-of-part-of-the-curve-with-equation--y-=-4x³-+-9x²---30x---8,---0.5-≤-x-≤-2.2--The-curve-has-a-turning-point-at-the-point-A-Edexcel-A-Level Maths Pure-Question 1-2016-Paper 3.png

Figure 2 shows a sketch of part of the curve with equation y = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A. (a) Using calcul... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation y = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 1 - 2016 - Paper 3

Step 1

Using calculus, show that the x coordinate of A is 1.

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Answer

To find the turning point of the curve, we first need to differentiate the equation:

$\frac{dy}{dx} = 12x^2 + 18x - 30$

Next, we set the derivative equal to zero to find the critical points:

$12x^2 + 18x - 30 = 0$

Using the quadratic formula where ( a = 12 ), ( b = 18 ), ( c = -30 ):

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-18 \pm \sqrt{(18)^2 - 4(12)(-30)}}{2(12)}$

Calculating within the square root:

$= \frac{-18 \pm \sqrt{324 + 1440}}{24} = \frac{-18 \pm \sqrt{1764}}{24} = \frac{-18 \pm 42}{24}$

Thus, we have:

( x = \frac{24}{24} = 1 )
( x = \frac{-60}{24} = -2.5 ) (not in range)

So, the x coordinate of A is indeed ( x = 1 ).

Step 2

Use integration to find the finite region R, giving your answer to 2 decimal places.

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Answer

To calculate the area of the region R, we need to find the area between the curve and the x-axis from the point B(2, 0) to point C(-\frac{1}{4}, 0).

The area A can be found using the integral:

$A = \int_{-\frac{1}{4}}^{2} (4x^3 + 9x^2 - 30x - 8) \, dx$

Calculating the definite integral:

Find the antiderivative:

$\int (4x^3 + 9x^2 - 30x - 8) \, dx = x^4 + 3x^3 - 15x^2 - 8x + C$

Evaluate from -\frac{1}{4} to 2:

$A = \left[ x^4 + 3x^3 - 15x^2 - 8x \right]_{-\frac{1}{4}}^{2}$

Calculating at the upper limit (x = 2):

$= 2^4 + 3(2)^3 - 15(2)^2 - 8(2) = 16 + 24 - 60 - 16 = -36$

Calculating at the lower limit (x = -\frac{1}{4}):

$= \left(-\frac{1}{4}\right)^4 + 3\left(-\frac{1}{4}\right)^3 - 15\left(-\frac{1}{4}\right)^2 - 8\left(-\frac{1}{4}\right)$

Calculating each term:

(= \frac{1}{256} )
(= -\frac{3}{64} )
(= -\frac{15}{16} = -\frac{240}{256} )
(= 2 )

So this becomes:

$= \frac{1}{256} - \frac{3}{64} - \frac{240}{256} + 2 = \frac{1 - 12 - 240 + 512}{256} = \frac{261}{256}$

Therefore, the area A is:

$A = -36 - \left(\frac{261}{256}\right) = -36 + \frac{261}{256} = \frac{-9216 + 261}{256} = \frac{-8955}{256}$

To two decimal places, approximately:

$A \approx 32.52$

Hence, the area of the finite region R is (32.52).

Figure 2 shows a sketch of part of the curve with equation y = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 1 - 2016 - Paper 3

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation y = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 1 - 2016 - Paper 3

Using calculus, show that the x coordinate of A is 1.

Use integration to find the finite region R, giving your answer to 2 decimal places.

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