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Below is a problem I did. I believe I got the right answer. However, I am not convinced my method is correct. I am hoping somebody here can verify that my solution is correct or tell me where I am wrong.

Problem:

Evaluate the following integral by evaluating its limit as a Riemann sum. $$ \int_0^1 3x^2 \mathrm{d}x $$

Answer:

Here is the definition of a Riemann sum, $R$. $$ R = \sum_{i = 1}^{n} f( x_i ) \Delta x $$ where $ x_i = a + i \Delta x $ and $\Delta x = \frac{b-a}{n} $ Now we have: \begin{align*} a &= 0 \\ b &= 1 \\ \Delta x &= \frac{1 - 0}{n} = \frac{1}{n} \\ \int_0^1 3x^2 \mathrm{d}x &= \lim_{n \rightarrow \infty } \sum_{i = 1}^{n} f( x_i ) \Delta x \\ &= \lim_{n \rightarrow \infty } \sum_{i = 1}^{n} \frac{ f( x_i ) }{ n } \\ &= \lim_{n \rightarrow \infty } \sum_{i = 1}^{n} \frac{ 3 x_i^2 }{ n } \\ &= \lim_{n \rightarrow \infty } \sum_{i = 1}^{n} \frac{ 3(a + i \Delta x)^2 }{ n }\\ &= \lim_{n \rightarrow \infty } \sum_{i = 1}^{n} \frac{ 3(0 + i \Delta x)^2 }{ n } \\ &= \lim_{n \rightarrow \infty } \sum_{i = 1}^{n} \frac{ 3( i^2 \left( \Delta x\right) ^2 ) }{ n } \\ &= \lim_{n \rightarrow \infty } \sum_{i = 1}^{n} \frac{ 3 i^2 }{ n^3 } \\ &= 3 \lim_{n \rightarrow \infty } \sum_{i = 1}^{n} \frac{ i^2 }{ n^3 } \\ \end{align*} Recall that: $$ \sum_{i = 1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} $$ $$\lim_{n \rightarrow \infty } \sum_{i = 1}^{n} f( x_i ) \Delta x = 3 \lim_{n \rightarrow \infty } \frac{ n(n+1)(2n+1) }{ 6n^3 } = \frac{3(2)}{6} = 1.$$

To check my answer, I integrate the function. $$ \left. \int_0^1 3x^2 \mathrm{d}x = x^3 \right|_0^1 = 1 - 0 = 1.$$ Therefore, my answer checks.

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    $\begingroup$ It seems correct to me! $\endgroup$
    – Physor
    Commented Oct 1, 2020 at 0:19
  • $\begingroup$ This is not a worthwhile question. You believe you have got the correct answer. You are unsure if your method is correct. But you don't provide any reason for doubting it. The fact that you got the correct answer strongly suggests that the method is correct. For such a question to be worthwhile you must provide a good reason for doubting something which is apparently correct. $\endgroup$ Commented Oct 1, 2020 at 0:58

1 Answer 1

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I know believe that my answer is correct.

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