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4. (a) Express $$\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} \delta x \cdot \frac{2}{x}$$ as an integral - Edexcel - A-Level Maths Pure - Question 4 - 2022 - Paper 1

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4.-(a)-Express-$$\lim_{\alpha-\to-0}-\sum_{x=2.1}^{6.3}-\delta-x-\cdot-\frac{2}{x}$$-as-an-integral-Edexcel-A-Level Maths Pure-Question 4-2022-Paper 1.png

4. (a) Express $$\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} \delta x \cdot \frac{2}{x}$$ as an integral. (b) Hence show that $$\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} \d... show full transcript

Worked Solution & Example Answer:4. (a) Express $$\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} \delta x \cdot \frac{2}{x}$$ as an integral - Edexcel - A-Level Maths Pure - Question 4 - 2022 - Paper 1

Step 1

Express $$\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} \delta x \cdot \frac{2}{x}$$ as an integral.

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Answer

To express the limit as an integral, we recognize that the summation is essentially a Riemann sum. Therefore, we can express:

limα0x=2.16.3δx2x=2.16.32xdx.\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} \delta x \cdot \frac{2}{x} = \int_{2.1}^{6.3} \frac{2}{x} \, dx.

Step 2

Hence show that $$\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} \delta x \cdot \frac{2}{x} = \ln k$$ where $k$ is a constant to be found.

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Answer

From the previous part, we have:

2.16.32xdx.\int_{2.1}^{6.3} \frac{2}{x} \, dx.

Now, we calculate the integral:

2xdx=2lnx+C,\int \frac{2}{x} \, dx = 2 \ln |x| + C,

Thus, we can evaluate the definite integral:

2ln(x)2.16.3=2ln(6.3)2ln(2.1)=2[ln(6.3)ln(2.1)]=2ln(6.32.1).2 \ln(x) \Big|_{2.1}^{6.3} = 2 \ln(6.3) - 2 \ln(2.1) = 2 [\ln(6.3) - \ln(2.1)] = 2 \ln\left( \frac{6.3}{2.1} \right).

We set this equal to lnk\ln k, which simplifies to:

ln(k)=2ln(3)\ln\left( k \right) = 2 \ln\left( 3 \right) leading to:

k=e2ln3=32=9.k = e^{2 \ln 3} = 3^2 = 9.

Thus, we have shown that:

limα0x=2.16.3δx2x=ln9.\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} \delta x \cdot \frac{2}{x} = \ln 9.

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