Figure 4 shows a sketch of the curve C with equation y = 5x^2 - 9x + 11, x > 0 The point P with coordinates (4, 15) lies on C. The line l is the tangent to C at the point P. The region R, shown shaded in Figure 4, is bounded by the curve C, the line l and the y-axis. Show that the area of R is 24, making your method clear. (Solutions based entirely on graphical or numerical methods are not acceptable.)

A-Level Edexcel Maths: Pure Integration: Figure 4 shows a sketch of the c

Figure 4 shows a sketch of the curve C with equation y = 5x^2 - 9x + 11, x > 0 The point P with coordinates (4, 15) lies on C - Edexcel - A-Level Maths: Pure - Question 1 - 2017 - Paper 2

Question 1

Figure 4 shows a sketch of the curve C with equation y = 5x^2 - 9x + 11, x > 0 The point P with coordinates (4, 15) lies on C. The line l is the tangent to C at t... show full transcript

Worked Solution & Example Answer:Figure 4 shows a sketch of the curve C with equation y = 5x^2 - 9x + 11, x > 0 The point P with coordinates (4, 15) lies on C - Edexcel - A-Level Maths: Pure - Question 1 - 2017 - Paper 2

Step 1

Differentiate the function

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Answer

To find the gradient of the curve C, we differentiate the given equation:

$\frac{dy}{dx} = 10x - 9$

We can evaluate this at the point P to find the gradient:

Step 2

Evaluate the gradient at P

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Answer

Substituting $x = 4$ into the derivative gives us:

$\frac{dy}{dx}\big|_{x=4} = 10(4) - 9 = 40 - 9 = 31$

Thus, the gradient at point P (4, 15) is 31.

Step 3

Find the equation of the tangent line l

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Answer

Using the point-slope form of the equation of a line, we find:

$y - y_1 = m(x - x_1)$

Substituting $(x_1, y_1) = (4, 15)$ and $m = 31$ gives:

$y - 15 = 31(x - 4)$ $y = 31x - 124 + 15$ $$y = 31x - 109$$$$

Step 4

Set up the integral to find the area R

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Answer

The area of region R is bounded between the curve C and the line l from $x = 0$ to $x = 4$ . Thus:

$\text{Area} = \int_0^4 \left( (5x^2 - 9x + 11) - (31x - 109) \right) dx$

Step 5

Simplify the integrand and compute the integral

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Answer

The integrand simplifies to:

$5x^2 - 9x + 11 - 31x + 109 = 5x^2 - 40x + 120$

Now, compute the integral:

$\int_0^4 (5x^2 - 40x + 120) dx$

Step 6

Evaluate the integral

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Answer

Calculating the integral yields:

$\left[ \frac{5}{3}x^3 - 20x^2 + 120x \right]_0^4$

Evaluating the bounds gives:

$= \left( \frac{5}{3} (4^3) - 20 (4^2) + 120(4) \right) - 0$ $= \left( \frac{5}{3} \cdot 64 - 320 + 480 \right)$ $= \frac{320}{3} - 320 + 480$ $= \frac{320 - 960 + 1440}{3} = \frac{800}{3}$

After simplification, we find:

$\text{Area} = 24$

Figure 4 shows a sketch of the curve C with equation y = 5x^2 - 9x + 11, x > 0 The point P with coordinates (4, 15) lies on C - Edexcel - A-Level Maths: Pure - Question 1 - 2017 - Paper 2

Worked Solution & Example Answer:Figure 4 shows a sketch of the curve C with equation y = 5x^2 - 9x + 11, x > 0 The point P with coordinates (4, 15) lies on C - Edexcel - A-Level Maths: Pure - Question 1 - 2017 - Paper 2

Differentiate the function

Evaluate the gradient at P

Find the equation of the tangent line l

Set up the integral to find the area R

Simplify the integrand and compute the integral

Evaluate the integral

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