Figure 3 shows a sketch of part of the curve with equation y = 7x^2(5 - 2√x), during x > 0. The curve has a turning point at the point A, where x > 0, as shown in Figure 3. (a) Using calculus, find the coordinates of the point A. The curve crosses the x-axis at the point B, as shown in Figure 3. (b) Use algebra to find the x coordinate of the point B. (c) The finite region R, shown shaded in Figure 3, is bounded by the curve, the line through A parallel to the x-axis and the line through B parallel to the y-axis. Use integration to find the area of the region R, giving your answer to 2 decimal places.

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Figure 3 shows a sketch of part of the curve with equation y = 7x^2(5 - 2√x), during x > 0 - Edexcel - A-Level Maths Pure - Question 2 - 2018 - Paper 1

Question 2

Figure 3 shows a sketch of part of the curve with equation y = 7x^2(5 - 2√x), during x > 0. The curve has a turning point at the point A, where x > 0, as shown in... show full transcript

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation y = 7x^2(5 - 2√x), during x > 0 - Edexcel - A-Level Maths Pure - Question 2 - 2018 - Paper 1

Step 1

(a) Using calculus, find the coordinates of the point A.

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Answer

To find the coordinates of point A, we begin by differentiating the equation of the curve:

$\frac{dy}{dx} = 70x - 35x^{\frac{3}{2}}$

Set the derivative equal to zero to find the turning points:

$70x - 35x^{\frac{3}{2}} = 0$

Factoring gives:

$35x(2 - \sqrt{x}) = 0$

This yields two solutions: $x = 0$ or $\sqrt{x} = 2$ , leading to $x = 4$ . Now substituting $x = 4$ back into the original equation to find y:

$y = 7(4^2)(5 - 2\sqrt{4}) = 7(16)(5 - 4) = 7(16)(1) = 112$

Thus, the coordinates of point A are (4, 112).

Step 2

(b) Use algebra to find the x coordinate of the point B.

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Answer

To find the x-coordinate at point B where the curve crosses the x-axis, set y to zero:

$0 = 7x^2(5 - 2\sqrt{x})$

This gives us two cases: $7x^2 = 0$ (which implies $x = 0$ , not in consideration) or $5 - 2\sqrt{x} = 0$ .

From the latter:

$2\sqrt{x} = 5$ $\sqrt{x} = \frac{5}{2}$

Squaring both sides:

$x = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25$

Thus, the x-coordinate of point B is 6.25.

Step 3

(c) Use integration to find the area of the region R, giving your answer to 2 decimal places.

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Answer

To find the area of region R between points A and B, we need to calculate:

$\int_{4}^{6.25} (7x^2(5 - 2\sqrt{x})) \, dx$

This can be expanded and computed separately:

First, compute the integral:

$= \int_{4}^{6.25}(35x^2 - 14x^{\frac{5}{2}}) \, dx$
Integrate each term:

$= \left[\frac{35x^3}{3} - \frac{14x^{\frac{7}{2}}}{\frac{7}{2}}\right]_{4}^{6.25}$
Evaluating:

At x = 6.25:

$\frac{35(6.25)^3}{3} - 4(6.25)^{\frac{7}{2}}$

At x = 4:

$\frac{35(4^3)}{3} - 4(4^{\frac{7}{2}})$
Calculate the definite integral and simplify:

This will yield the total area in the form of a decimal:

Giving the area as approximately 172.23 after calculation, rounded appropriately gives the final answer of: 172.23.

Figure 3 shows a sketch of part of the curve with equation y = 7x^2(5 - 2√x), during x > 0 - Edexcel - A-Level Maths Pure - Question 2 - 2018 - Paper 1

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation y = 7x^2(5 - 2√x), during x > 0 - Edexcel - A-Level Maths Pure - Question 2 - 2018 - Paper 1

(a) Using calculus, find the coordinates of the point A.

(b) Use algebra to find the x coordinate of the point B.

(c) Use integration to find the area of the region R, giving your answer to 2 decimal places.

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