Figure 1 shows the graph of the curve with equation $$y = xe^{-x^2}, \quad x \geq 0.$$ The finite region $R$ bounded by the lines $x = 1$, the $x$-axis and the curve is shown shaded in Figure 1. (a) Use integration to find the exact value of the area for $R$. (b) Complete the table with the values of $y$ corresponding to $x = 0.4$ and $0.8$. (c) Use the trapezium rule with all the values in the table to find an approximate value for this area, giving your answer to 4 significant figures.

A-Level Edexcel Maths Pure Vectors in 2 Dimensions: Figure 1 shows the graph of the

Figure 1 shows the graph of the curve with equation $$y = xe^{-x^2}, \quad x \geq 0.$$ The finite region $R$ bounded by the lines $x = 1$, the $x$-axis and the curve is shown shaded in Figure 1 - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 6

Question 7

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Figure 1 shows the graph of the curve with equation $$y = xe^{-x^2}, \quad x \geq 0.$$ The finite region $R$ bounded by the lines $x = 1$, the $x$-axis and the c... show full transcript

Worked Solution & Example Answer:Figure 1 shows the graph of the curve with equation $$y = xe^{-x^2}, \quad x \geq 0.$$ The finite region $R$ bounded by the lines $x = 1$, the $x$-axis and the curve is shown shaded in Figure 1 - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 6

Step 1

Use integration to find the exact value of the area for $R$

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Answer

To find the exact area of the region $R$ , we need to calculate the integral:

$A = \int_0^1 xe^{-x^2} \, dx.$

We can use the substitution method. Let:

$u = -x^2 \Rightarrow du = -2x \, dx \Rightarrow dx = -\frac{du}{2x}.$

When $x = 0$ , $u = 0$ and when $x = 1$ , $u = -1$ . Thus,

$A = \int_0^{-1} e^u \left(-\frac{1}{2} \right) du \quad = \frac{1}{2} \int_{-1}^0 e^u \, du \quad = \frac{1}{2} \, \left[ e^u \right]_{-1}^0 = \frac{1}{2} \left(e^0 - e^{-1} \right) = \frac{1}{2} \left(1 - \frac{1}{e}\right).$

Thus, the exact area for $R$ is approximately:

$A \approx \frac{1}{2} \left(1 - \frac{1}{e}\right).$

Step 2

Complete the table with the values of $y$ corresponding to $x = 0.4$ and $0.8$

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Answer

To compute the values of $y$ corresponding to $x = 0.4$ and $x = 0.8$ , we substitute these values into the equation of the curve:

For $x = 0.4$ :

$y = 0.4 e^{-0.4^2} \approx 0.29836.$
For $x = 0.8$ :

$y = 0.8 e^{-0.8^2} \approx 1.99207.$

Thus, the completed values are:

$x$	$y$
0	0
0.2	0.29836
0.4	0.8
0.6	1.99207
0.8	1.99207
1	7.38906

Step 3

Use the trapezium rule with all the values in the table to find an approximate value for this area

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Answer

Using the trapezium rule for the values calculated in the table, we can estimate the area as follows:

$I \approx \frac{h}{2}(y_0 + 2y_1 + 2y_2 + ... + 2y_{n-1} + y_n)$

where $h = 0.2$ , and $y_0 = 0$ , $y_1 = 0.29836$ , $y_2 = 1.99207$ , $y_3 = 7.38906$ .

Calculating this gives:

I & \approx \frac{0.2}{2}(0 + 2(0.29836 + 1.99207) + 7.38906) \\ & \approx 0.1(0 + 2(2.29043) + 7.38906) \\ & \approx 0.1(4.58086 + 7.38906) \\ & \approx 0.1(11.96992) \\ & \approx 1.196992. \\ & \text{Therefore, the approximate value of the area is:} \approx 1.1970. \end{align*}$$

Figure 1 shows the graph of the curve with equation $$y = xe^{-x^2}, \quad x \geq 0.$$ The finite region $R$ bounded by the lines $x = 1$, the $x$-axis and the curve is shown shaded in Figure 1 - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 6

Worked Solution & Example Answer:Figure 1 shows the graph of the curve with equation $$y = xe^{-x^2}, \quad x \geq 0.$$ The finite region $R$ bounded by the lines $x = 1$, the $x$-axis and the curve is shown shaded in Figure 1 - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 6

Use integration to find the exact value of the area for $R$

Complete the table with the values of $y$ corresponding to $x = 0.4$ and $0.8$

Use the trapezium rule with all the values in the table to find an approximate value for this area

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