Figure 1 shows a sketch of part of the curve with equation $$y = \frac{(x + 2)^{\frac{3}{2}}}{4}$$ where $$x > -2$$. The finite region $$R$$, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line with equation $$x = 10$$. The table below shows corresponding values of $$x$$ and $$y$$ for $$y = \frac{(x + 2)^{\frac{3}{2}}}{4}$$. (a) Complete the table, giving values of $$y$$ corresponding to $$x = 2$$ and $$x = 6$$. | $$x$$ | $$-2$$ | $$2$$ | $$6$$ | $$10$$ | |-------|-------|-------|-------|-------| | $$y$$ | $$0$$ | $$2$$ | $$4√2$$ | $$6√3$$ | (b) Use the trapezium rule, with all the values of $$y$$ from the completed table, to find an approximate value for the area of $$R$$, giving your answer to 3 decimal places.

A-Level Edexcel Maths Pure Polynomials: Figure 1 shows a sketch of part

Figure 1 shows a sketch of part of the curve with equation $$y = \frac{(x + 2)^{\frac{3}{2}}}{4}$$ where $$x > -2$$ - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 4

Question 3

$Figure-1-shows-a-sketch-of-part-of-the-curve-with-equation--$$y-=-\frac{(x-+-2)^{\frac{3}{2}}}{4}$$--where-$$x->--2$$-Edexcel-A-Level Maths Pure-Question 3-2018-Paper 4.png$

Figure 1 shows a sketch of part of the curve with equation $$y = \frac{(x + 2)^{\frac{3}{2}}}{4}$$ where $$x > -2$$. The finite region $$R$$, shown shaded in Figu... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of part of the curve with equation $$y = \frac{(x + 2)^{\frac{3}{2}}}{4}$$ where $$x > -2$$ - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 4

Step 1

Complete the table, giving values of y corresponding to x = 2 and x = 6

96%

114 rated

Answer

To find the values of $y$ corresponding to the given values of $x$ , we will plug in the values into the equation:

For $x = 2$ :

$y = \frac{(2 + 2)^{\frac{3}{2}}}{4} = \frac{(4)^{\frac{3}{2}}}{4} = \frac{8}{4} = 2$ .
For $x = 6$ :

$y = \frac{(6 + 2)^{\frac{3}{2}}}{4} = \frac{(8)^{\frac{3}{2}}}{4} = \frac{16√2}{4} = 4√2$ .

Thus, the completed table is:

$x$	$-2$	$2$	$6$	$10$
$y$	$0$	$2$	$4√2$	$6√3$

Step 2

Use the trapezium rule, with all the values of y from the completed table, to find an approximate value for the area of R

99%

104 rated

Answer

To calculate the area of region $R$ , we will use the trapezium rule with the values obtained from the table.

The area $A$ is given by the formula:

$A = \frac{1}{2}h \times (y_0 + 2y_1 + 2y_2 + y_n)$

Where:

$h$ is the width of each segment (step size), which is $2$ for this case
$y_0 = 0$ , $y_1 = 2$ , $y_2 = 4√2$ , $y_3 = 6√3$

This gives us:

The width $h = 2$ for each interval.
Therefore:

$A = \frac{1}{2} \times 2 \times (0 + 2(2) + 2(4√2) + 6√3)$ $A = 1 (0 + 4 + 8√2 + 6√3)$ $A = 4 + 8√2 + 6√3$

Calculating:

Approximate $8√2 \approx 11.3137$
$6√3 \approx 10.3923$

Thus:

$A \approx 4 + 11.3137 + 10.3923 \approx 25.706$

Rounding to 3 decimal places:

$A \approx 25.706$ .

Figure 1 shows a sketch of part of the curve with equation $$y = \frac{(x + 2)^{\frac{3}{2}}}{4}$$ where $$x > -2$$ - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 4

Worked Solution & Example Answer:Figure 1 shows a sketch of part of the curve with equation $$y = \frac{(x + 2)^{\frac{3}{2}}}{4}$$ where $$x > -2$$ - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 4

Complete the table, giving values of y corresponding to x = 2 and x = 6

Use the trapezium rule, with all the values of y from the completed table, to find an approximate value for the area of R

Join the A-Level students using SimpleStudy...