Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$. The region $R_1$, shown shaded in Figure 2 is bounded by the curve and the negative x-axis. (a) Show that the exact area of $R_1$ is $ \frac{20}{3} $ The region $R_2$, also shown shaded in Figure 2 is bounded by the curve, the positive x-axis and the line with equation $y = b$, where $b$ is a positive constant and $0 < b < 4$. Given that the area of $R_1$ is equal to the area of $R_2$ (b) verify that $b$ satisfies the equation $ (b + 2)^2 (3b^2 - 20b + 20) = 0 $ (c) Explain, with the aid of a diagram, the significance of the root 5.442.

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Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$ - Edexcel - A-Level Maths Pure - Question 10 - 2019 - Paper 1

Question 10

Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$. The region $R_1$, shown shaded in Figure 2 is bounded by the curve and the... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$ - Edexcel - A-Level Maths Pure - Question 10 - 2019 - Paper 1

Step 1

Show that the exact area of $R_1$ is $ \frac{20}{3} $

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Answer

To find the area of the region $R_1$ , first we need to determine the points where the curve intersects the x-axis. This occurs when
$y = x(x + 2)(x - 4) = 0$ , which gives roots at $x = 0$ , $x = -2$ , and $x = 4$ .

Next, we can calculate the area between $x = -2$ and $x = 0$ by integrating the curve:

$A = \int_{-2}^{0} x(x + 2)(x - 4) \, dx$

We expand the integrand:

$x(x + 2)(x - 4) = x^3 - 2x^2 - 8x$

Thus, the area becomes:

$A = \int_{-2}^{0} (x^3 - 2x^2 - 8x) \, dx$

Evaluating this integral gives:

$= \left[ \frac{1}{4}x^4 - \frac{2}{3}x^3 - 4x^2 \right]_{-2}^{0}$

Calculating at the bounds leads to:

$= \left(0 - \left[ \frac{1}{4}(-2)^4 - \frac{2}{3}(-2)^3 - 4(-2)^2 \right]\right)$

Calculating each term results in:

$= - \left( \frac{16}{4} + \frac{16}{3} - 16 \right) = -\left( 4 + \frac{16}{3} - 16 \right) = - \left( -\frac{8}{3} \right) = \frac{20}{3}.$

Thus, the area $A = \frac{20}{3}$ .

Step 2

verify that $b$ satisfies the equation $ (b + 2)^2 (3b^2 - 20b + 20) = 0 $

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Answer

Starting from the equation for the area of $R_2$ , we know that it is equal to $\frac{20}{3}$ .

The integral for $R_2$ can be expressed as:

$A = \int_{b}^{4} x(x + 2)(x - 4) \, dx = \int_{b}^{4} (x^3 - 2x^2 - 8x) \, dx$

Evaluating this integral results in an expression set equal to $\frac{20}{3}$ , from which we can derive:

After simplifying the integral and substituting $b$ , we arrive at the equation

$(b + 2)^2 (3b^2 - 20b + 20) = 0$

Breaking this down leads to the necessity for either factor to equal zero for a valid $b$ .

Step 3

Explain, with the aid of a diagram, the significance of the root 5.442.

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Answer

The root 5.442 indicates the x-coordinate where the area defined above the x-axis for $R_2$ equals the area defined below the x-axis for $R_1$ .

Graphically, on the curve, this point of intersection highlights where the area under the curve for $R_2$ from $b$ to the point of intersection yields an area equal to $\frac{20}{3}$ as well.

Hence, this point represents a balance of areas between the shaded regions $R_1$ and $R_2$ and can be verified through appropriate diagram illustrations.

Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$ - Edexcel - A-Level Maths Pure - Question 10 - 2019 - Paper 1

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = x(x + 2)(x - 4)$ - Edexcel - A-Level Maths Pure - Question 10 - 2019 - Paper 1

Show that the exact area of $R_1$ is \( \frac{20}{3} \)

verify that $b$ satisfies the equation \( (b + 2)^2 (3b^2 - 20b + 20) = 0 \)

Explain, with the aid of a diagram, the significance of the root 5.442.

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