A-Level AQA Maths Pure Modelling involving Numerical Methods: 6 (a) Find the first three terms

6 (a) Find the first three terms, in ascending powers of x, of the binomial expansion of $$\frac{1}{\sqrt{4+x}}$$ 6 (b) Hence, find the first three terms of the binomial expansion of $$\frac{1}{\sqrt{4 - x^3}}$$ 6 (c) Using your answer to part (b), find an approximation for $$\int_0^1 \frac{1}{\sqrt{4 - x^3}} \mathrm{d}x$$, giving your answer to seven decimal places - AQA - A-Level Maths Pure - Question 6 - 2018 - Paper 1

Question 6

$6-(a)-Find-the-first-three-terms,-in-ascending-powers-of-x,-of-the-binomial-expansion-of--$$\frac{1}{\sqrt{4+x}}$$--6-(b)-Hence,-find-the-first-three-terms-of-the-binomial-expansion-of--$$\frac{1}{\sqrt{4---x^3}}$$--6-(c)-Using-your-answer-to-part-(b),-find-an-approximation-for--$$\int_0^1-\frac{1}{\sqrt{4---x^3}}-\mathrm{d}x$$,-giving-your-answer-to-seven-decimal-places-AQA-A-Level Maths Pure-Question 6-2018-Paper 1.png$

6 (a) Find the first three terms, in ascending powers of x, of the binomial expansion of $$\frac{1}{\sqrt{4+x}}$$ 6 (b) Hence, find the first three terms of the bi... show full transcript

Worked Solution & Example Answer:6 (a) Find the first three terms, in ascending powers of x, of the binomial expansion of $$\frac{1}{\sqrt{4+x}}$$ 6 (b) Hence, find the first three terms of the binomial expansion of $$\frac{1}{\sqrt{4 - x^3}}$$ 6 (c) Using your answer to part (b), find an approximation for $$\int_0^1 \frac{1}{\sqrt{4 - x^3}} \mathrm{d}x$$, giving your answer to seven decimal places - AQA - A-Level Maths Pure - Question 6 - 2018 - Paper 1

Step 1

Find the first three terms, in ascending powers of x, of the binomial expansion of $$\frac{1}{\sqrt{4+x}}$$

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Answer

To apply the binomial expansion, we first rewrite the term:

$\frac{1}{\sqrt{4+x}} = (4+x)^{-\frac{1}{2}}$ .

Using the binomial series formula, we get:

$(1 + k)^{n} = 1 + nk + \frac{n(n-1)}{2!}k^2 + ...$

Letting k = \frac{x}{4} and n = -\frac{1}{2}$, the first three terms are:

The first term: $1$ .
The second term: $-\frac{1}{2} \cdot \frac{x}{4} = -\frac{x}{8}$ .
The third term: $\frac{-\frac{1}{2} \cdot -\frac{3}{2}}{2} \cdot \left(\frac{x}{4}\right)^2 = \frac{3x^2}{128}$ .

Thus, the first three terms are:

$1 - \frac{x}{8} + \frac{3x^2}{128}$ .

Step 2

Hence, find the first three terms of the binomial expansion of $$\frac{1}{\sqrt{4 - x^3}}$$

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Answer

For this expression, we use similar steps.

We rewrite it as:

$\frac{1}{\sqrt{4(1 - \frac{x^3}{4})}} = \frac{1}{2\sqrt{1 - \frac{x^3}{4}}}$ .

Applying the binomial expansion:

First term: $\frac{1}{2}$ .
Second term: $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{x^3}{4} = \frac{x^3}{16}$ .
The third term: $\frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{-3}{2} \cdot \left(\frac{x^3}{4}\right)^2 = -\frac{3x^6}{128}$ .

The first three terms are:

$\frac{1}{2} + \frac{x^3}{16} - \frac{3x^6}{128}$ .

Step 3

Using your answer to part (b), find an approximation for $$\int_0^1 \frac{1}{\sqrt{4 - x^3}} \mathrm{d}x$$, giving your answer to seven decimal places.

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Answer

Using the expansion from part (b), we approximate:

$\int_0^1 \left( \frac{1}{2} + \frac{x^3}{16} - \frac{3x^6}{128} \right) \mathrm{d}x$ .

Calculating each term:

For $\frac{1}{2}$ : $\int_0^1 \frac{1}{2} \mathrm{d}x = \frac{1}{2}.$
For $\frac{x^3}{16}$ : $\int_0^1 \frac{x^3}{16} \mathrm{d}x = \frac{1}{16} \cdot \frac{1}{4} = \frac{1}{64}.$
For $-\frac{3x^6}{128}$ : $\int_0^1 -\frac{3x^6}{128} \mathrm{d}x = -\frac{3}{128} \cdot \frac{1}{7} = -\frac{3}{896}.$

Combining all these, we find: $$\frac{1}{2} + \frac{1}{64} - \frac{3}{896} = \frac{448}{896} + \frac{14}{896} - \frac{3}{896} = \frac{459}{896} \approx 0.5127991.$

Step 4

Edward, a student, decides to use this method to find a more accurate value for the integral by increasing the number of terms of the binomial expansion used. Explain clearly whether Edward's approximation will be an overestimate, an underestimate, or if it is impossible to tell.

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Answer

Each term in the expansion is positive, so increasing the number of terms will always yield a more accurate estimate of the integral. Thus, Edward's approximation will be an underestimate.

Step 5

Edward goes on to use the expansion from part (b) to find an approximation for $$\int_0^{\frac{1}{2}} \frac{1}{\sqrt{4 - x^3}} \mathrm{d}x$$. Explain why Edward's approximation is invalid.

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Answer

The validity of the binomial expansion is contingent upon the term $|x| < \sqrt{4}$ . When Edward uses the value $\frac{1}{2}$ , it is less than $\\sqrt{4}$ , hence valid according to the range. However, as he expands the series, the terms become overly simplified which violates the original expression, resulting in a loss of accuracy in the approximation. Thus, the approximation is invalid.

Find the first three terms, in ascending powers of x, of the binomial expansion of $$\frac{1}{\sqrt{4+x}}$$

Hence, find the first three terms of the binomial expansion of $$\frac{1}{\sqrt{4 - x^3}}$$

Using your answer to part (b), find an approximation for $$\int_0^1 \frac{1}{\sqrt{4 - x^3}} \mathrm{d}x$$, giving your answer to seven decimal places.

Edward, a student, decides to use this method to find a more accurate value for the integral by increasing the number of terms of the binomial expansion used. Explain clearly whether Edward's approximation will be an overestimate, an underestimate, or if it is impossible to tell.

Edward goes on to use the expansion from part (b) to find an approximation for $$\int_0^{\frac{1}{2}} \frac{1}{\sqrt{4 - x^3}} \mathrm{d}x$$. Explain why Edward's approximation is invalid.

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