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.
Best Answer
I know believe that my answer is correct.