The table below shows corresponding values of x and y for $y = \frac{x}{\sqrt{1+x}}$. The values of y are given to 4 significant figures. | x | y | |-----|------| | 0.5 | 0.5774 | | 1 | 0.7071 | | 1.5 | 0.7746 | | 2 | 0.8165 | | 2.5 | 0.8452 | (a) Use the trapezium rule, with all the values of y in the table, to find an estimate for $$\int_{0.5}^{2.5} \frac{x}{1+x} \, dx$$ giving your answer to 3 significant figures. (b) Using your answer to part (a), deduce an estimate for $$\int_{0.5}^{2.5} \frac{9x}{1+x} \, dx$$ (c) comment on the accuracy of your answer to part (b).

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The table below shows corresponding values of x and y for $y = \frac{x}{\sqrt{1+x}}$ - Edexcel - A-Level Maths Pure - Question 3 - 2020 - Paper 2

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The table below shows corresponding values of x and y for $y = \frac{x}{\sqrt{1+x}}$. The values of y are given to 4 significant figures. | x | y | |-----|---... show full transcript

Worked Solution & Example Answer:The table below shows corresponding values of x and y for $y = \frac{x}{\sqrt{1+x}}$ - Edexcel - A-Level Maths Pure - Question 3 - 2020 - Paper 2

Step 1

Use the trapezium rule to find an estimate for $$\int_{0.5}^{2.5} \frac{x}{1+x} \, dx$$

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Answer

To approximate the integral using the trapezium rule, we first identify the values of x and corresponding y from the table. The width of each sub-interval is given by:

$h = 0.5$

Next, we apply the trapezium rule formula:

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

Using the values from the table:

$y_0 = 0.5774$ (for x = 0.5)
$y_1 = 0.7071$ (for x = 1)
$y_2 = 0.7746$ (for x = 1.5)
$y_3 = 0.8165$ (for x = 2)
$y_4 = 0.8452$ (for x = 2.5)

We substitute these values into the formula:

$A = \frac{0.5}{2} \left(0.5774 + 2(0.7071 + 0.7746 + 0.8165) + 0.8452\right)$

Calculating further:

$A = 0.25 \left(0.5774 + 2(2.2982) + 0.8452\right)$

$= 0.25 \left(0.5774 + 4.5964 + 0.8452\right)$

$= 0.25 \times 6.019 \approx 1.50475$

Rounding to 3 significant figures gives: 1.50.

Step 2

Using your answer to part (a), deduce an estimate for $$\int_{0.5}^{2.5} \frac{9x}{1+x} \, dx$$

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Answer

From part (a), we found an estimate for the integral

$\int_{0.5}^{2.5} \frac{x}{1+x} \, dx \approx 1.50.$

To find the estimate for

$\int_{0.5}^{2.5} \frac{9x}{1+x} \, dx,$

we can multiply the result from part (a) by 9:

$\int_{0.5}^{2.5} \frac{9x}{1+x} \, dx \approx 9 \times 1.50 = 13.5.$

Thus, the estimate is 13.5.

Step 3

Comment on the accuracy of your answer to part (b)

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Answer

The accuracy of the estimate for the integral $\int_{0.5}^{2.5} \frac{9x}{1+x} \, dx$ depends on the trapezium rule's inherent approximation error.

Using trapezium rule estimations generally yields good results when the function is relatively linear over the intervals used. Since the function is smooth within the limits of integration, the estimate can be considered reasonably accurate.

However, due to the linear approximation, it could underestimate or overestimate the area under the curve. Furthermore, without conducting further tests or using numerical integration, it's challenging to ascertain the exact error margin.

Comparing the trapezium rule with exact integration, we get:

The exact integral can yield a more precise check against the approximation, indicating the possibility of refinement in the estimation methods.

The table below shows corresponding values of x and y for $y = \frac{x}{\sqrt{1+x}}$ - Edexcel - A-Level Maths Pure - Question 3 - 2020 - Paper 2

Worked Solution & Example Answer:The table below shows corresponding values of x and y for $y = \frac{x}{\sqrt{1+x}}$ - Edexcel - A-Level Maths Pure - Question 3 - 2020 - Paper 2

Use the trapezium rule to find an estimate for $$\int_{0.5}^{2.5} \frac{x}{1+x} \, dx$$

Using your answer to part (a), deduce an estimate for $$\int_{0.5}^{2.5} \frac{9x}{1+x} \, dx$$

Comment on the accuracy of your answer to part (b)

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