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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 4 - 2022 - Paper 1

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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 4-2022-Paper 1.png

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 = ... 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 4 - 2022 - Paper 1

Step 1

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

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Answer

To express the given limit as an integral, we begin by interpreting the summation. The expression can be related to the concept of definite integrals:

limα0x=2.16.32δx2.16.32dx\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x \approx \int_{2.1}^{6.3} 2 \, dx

In this case, we can see that as α\alpha approaches 0, the width of the partitions approaches 0, thus turning the Riemann sum into the integral above.

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.

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Answer

From the previous step, we have:

2.16.32dx\int_{2.1}^{6.3} 2 \, dx

Calculating this integral:

2dx=2x\int 2 \, dx = 2x

Evaluating from 2.1 to 6.3:

2[6.32.1]=2(6.32.1)=2(4.2)=8.42[6.3 - 2.1] = 2(6.3 - 2.1) = 2(4.2) = 8.4

Now, according to the marking scheme, we then need to equate this to lnk\ln k:

8.4=lnk8.4 = \ln k

To find kk, we exponentiate both sides:

k=e8.4k = e^{8.4}

Hence, we have shown that:

limα0x=2.16.32δx=ln(e8.4)=8.4\lim_{\alpha \to 0} \sum_{x=2.1}^{6.3} 2 \delta x = \ln(e^{8.4}) = 8.4, where k=e8.4k = e^{8.4}.

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