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4. (a) Express \( \lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x \) as an integral - Edexcel - A-Level Maths Pure - Question 6 - 2022 - Paper 1
Question 6
4. (a) Express \( \lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x \) as an integral. (b) Hence show that \( \lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x = \l... show full transcript
Worked Solution & Example Answer:4. (a) Express \( \lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x \) as an integral - Edexcel - A-Level Maths Pure - Question 6 - 2022 - Paper 1
Step 1
Express \( \lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x \) as an integral.
Answer
To express the limit of the sum as an integral, we recognize that [ \lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x = \int_{2.1}^{6.3} 2 , dx. ] This shows that the sum can be interpreted as the area under the curve of the function 2 over the interval from 2.1 to 6.3.
Step 2
Hence show that \( \lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x = \ln k \) where \( k \) is a constant to be found.
Answer
From the previous part, we have established that [ \lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x = \int_{2.1}^{6.3} 2 , dx = 2 (6.3 - 2.1) = 2 \cdot 4.2 = 8.4. ] Now, we set this equal to ( \ln k ), therefore: [ \ln k = 8.4. ] To find ( k ), we exponentiate both sides: [ k = e^{8.4}. ] Thus, we conclude that ( \lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x = \ln(e^{8.4}) = 8.4 ). Hence, within the context of this problem, the constant ( k = e^{8.4}.)