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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 $$ \int_{0}^{1} \frac{1}{\sqrt{4 - x^3}} dx $$, giving your answer to seven decimal places - AQA - A-Level Maths Mechanics - 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 t... 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 $$ \int_{0}^{1} \frac{1}{\sqrt{4 - x^3}} dx $$, giving your answer to seven decimal places - AQA - A-Level Maths Mechanics - Question 6 - 2018 - Paper 1
Step 1
Find the first three terms, in ascending powers of x, of the binomial expansion of \( \frac{1}{\sqrt{4 + x}} \)
Answer
To expand ( \frac{1}{\sqrt{4 + x}} ), we can re-write it as ( (4 + x)^{-\frac{1}{2}} ). Using the binomial expansion formula ( (1 + u)^n = 1 + nu + \frac{n(n - 1)}{2!}u^2 + \ldots ), where ( u = \frac{x}{4} ) and ( n = -\frac{1}{2} ):
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Substitute ( u = \frac{x}{4} ):
[ (1 + \frac{x}{4})^{-\frac{1}{2}} \approx 1 - \frac{1}{2}(\frac{x}{4}) + \frac{-\frac{1}{2} \cdot -\frac{3}{2}}{2!}(\frac{x}{4})^2 ] -
Calculate the first three terms:
- First term: ( 1 )
- Second term: ( -\frac{1}{8}x )
- Third term: ( +\frac{3}{256}x^2 )
Thus, the first three terms are ( 1 - \frac{1}{8}x + \frac{3}{256}x^2 ).
Step 2
Hence, find the first three terms of the binomial expansion of \( \frac{1}{\sqrt{4 - x^3}} \)
Answer
We can express ( \frac{1}{\sqrt{4 - x^3}} ) as ( (4 - x^3)^{-\frac{1}{2}} ). Similar to part (a), we can write it as:\n ( (4(1 - \frac{x^3}{4}))^{-\frac{1}{2}} )\nUsing the binomial expansion:
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Substitute ( u = -\frac{x^3}{4} ):
[ (1 - \frac{x^3}{4})^{-\frac{1}{2}} \approx 1 + \frac{1}{2}(\frac{x^3}{4}) + \frac{\frac{1}{2} \cdot \frac{3}{2}}{2!}(\frac{x^3}{4})^2 ] -
Calculate the first three terms:
- First term: ( 1 )
- Second term: ( +\frac{1}{8}x^3 )
- Third term: ( +\frac{3}{512}x^6 )
Thus, the first three terms are ( 1 + \frac{1}{8}x^3 + \frac{3}{512}x^6 ).
Step 3
Using your answer to part (b), find an approximation for \( \int_{0}^{1} \frac{1}{\sqrt{4 - x^3}} dx \)
Answer
Using the first three terms from part (b), we have:
[ \int_{0}^{1} (1 + \frac{1}{8}x^3 + \frac{3}{512}x^6) dx ]
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Integrate term by term:
- For ( 1 ): ( \int_{0}^{1} 1 , dx = 1 )
- For ( \frac{1}{8}x^3 ): ( \int_{0}^{1} \frac{1}{8}x^3 , dx = \frac{1}{8} \cdot \frac{x^4}{4} \bigg|_0^1 = \frac{1}{32} )
- For ( \frac{3}{512}x^6 ): ( \int_{0}^{1} \frac{3}{512}x^6 , dx = \frac{3}{512} \cdot \frac{x^7}{7} \bigg|_0^1 = \frac{3}{3584} )
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Combine the results:
[ 1 + \frac{1}{32} + \frac{3}{3584} \approx 1.0256504 ]
Thus, the approximation is ( 1.0256504 ).
Step 4
Explain clearly whether Edward's approximation will be an overestimate, an underestimate, or if it is impossible to tell.
Answer
Edward's approximation will lead to an underestimate. Since the binomial expansion is valid for values of (|x| < \sqrt{4}), the terms will continually approach the actual value but not exceed it. Therefore, the more terms added will refine the approximation, and as they are always positive in this case, higher order terms will increase the estimated area but never go over.
Step 5
State the validity of their binomial expansion for part (b), providing the reasoning.
Answer
The binomial expansion is valid for (|x| < \sqrt{4} = 2). This ensures that the expansion converges, allowing for accurate approximations within this range. Since we are evaluating for (x) within [0, 1], we remain well within the valid interval.