Figure 2 shows a sketch of part of the curve with equation $y = \ln(x)^2, \quad x > 0$ The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the line with equation $x = 2$, the x-axis and the line with equation $x = 4$. The table below shows corresponding values of $x$ and $y$, with the values of $y$ given to 4 decimal places. | x | 2 | 2.5 | 3 | 3.5 | 4 | |-----|-------|---------|-------|---------|-------| | y | 0.4805| 0.8396 | 1.2069| 1.5694 | 1.9218| (a) Use the trapezium rule, with all the values of $y$ in the table, to obtain an estimate for the area of $R$, giving your answer to 3 significant figures. (b) Use algebraic integration to find the exact area of $R$, giving your answer in the form $y = a(\ln(2))^2 + b\ln(2) + c$ where $a$, $b$ and $c$ are integers to be found.

A-Level Edexcel Maths Pure Circles: Figure 2 shows a sketch of part

Figure 2 shows a sketch of part of the curve with equation $y = \ln(x)^2, \quad x > 0$ The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the line with equation $x = 2$, the x-axis and the line with equation $x = 4$ - Edexcel - A-Level Maths Pure - Question 13 - 2021 - Paper 1

Question 13

$Figure-2-shows-a-sketch-of-part-of-the-curve-with-equation--$y-=-\ln(x)^2,-\quad-x->-0$--The-finite-region-$R$,-shown-shaded-in-Figure-2,-is-bounded-by-the-curve,-the-line-with-equation-$x-=-2$,-the-x-axis-and-the-line-with-equation-$x-=-4$-Edexcel-A-Level Maths Pure-Question 13-2021-Paper 1.png$

Figure 2 shows a sketch of part of the curve with equation $y = \ln(x)^2, \quad x > 0$ The finite region $R$, shown shaded in Figure 2, is bounded by the curve, th... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = \ln(x)^2, \quad x > 0$ The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the line with equation $x = 2$, the x-axis and the line with equation $x = 4$ - Edexcel - A-Level Maths Pure - Question 13 - 2021 - Paper 1

Step 1

Use the trapezium rule, with all the values of $y$ in the table, to obtain an estimate for the area of $R$, giving your answer to 3 significant figures.

96%

114 rated

Answer

To estimate the area of region $R$ using the trapezium rule, we apply the formula:

$A \approx \frac{h}{2} \left( y_0 + 2\sum_{i=1}^{n-1} y_i + y_n \right)$

where $h$ is the width of each interval. In this case, every interval is $0.5$ between $x=2$ and $x=4$ , giving us:

$h = 0.5$

Substituting values, we get:

$A \approx \frac{1}{2} \left( 0.4805 + 2(0.8396 + 1.2069 + 1.5694) + 1.9218 \right)$

Calculating the sum:

= \frac{1}{2} \left( 0.4805 + 2(3.6159) + 1.9218 \right)\ = \frac{1}{2} \left( 0.4805 + 7.2318 + 1.9218 \right)\ = \frac{1}{2} \left( 9.6341 \right) \ = 4.81705\ \approx 4.82 \text{ (to 3 significant figures)}

Step 2

Use algebraic integration to find the exact area of $R$, giving your answer in the form $y = a(\ln(2))^2 + b\ln(2) + c$.

99%

104 rated

Answer

To find the exact area of region $R$ , we integrate the function $y = \ln(x)^2$ from $x = 2$ to $x = 4$ :

$A = \int_{2}^{4} \ln(x)^2 dx$

Using integration by parts, let:

$u = \ln(x)^2 \Rightarrow du = \frac{2\ln(x)}{x} dx$
$dv = dx \Rightarrow v = x$

Then,

$A = \left[ x \ln(x)^2 \right]_{2}^{4} - \int_{2}^{4} x \cdot \frac{2\ln(x)}{x} dx$

This simplifies to:

$A = \left[ 4 \ln(4)^2 - 2 \cdot 2 \ln(2) \right] - 2\int_{2}^{4} \ln(x) dx$

Next, compute the second integral using the result of the first application of integration by parts:

$\int \ln(x) dx = x \ln(x) - x + C$

Thus, calculating:

$-2\left[ x \ln(x) - x \right]_{2}^{4} = -2\left[ (4\ln(4) - 4) - (2\ln(2) - 2) \right]$

After resolving these, substitute back into the area, and simplify to yield the final answer in the form:

$A = 4(\ln(4)^2 - 2 \ln(2) + 2)$ \text{which can be expressed as } A = 4(\ln(2)^2 + 2\ln(2) + c)$$

Adjust to match the required form for integers $a$ , $b$ , and $c$ .

Use the trapezium rule, with all the values of $y$ in the table, to obtain an estimate for the area of $R$, giving your answer to 3 significant figures.

Use algebraic integration to find the exact area of $R$, giving your answer in the form $y = a(\ln(2))^2 + b\ln(2) + c$.

Join the A-Level students using SimpleStudy...