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As the title says, I am trying to calculate the general formulas for the following integral. $$\int_0^1 (5+6x^2)dx$$

I've already completed a prior integral using the advice in this question, but I am having trouble following the same pattern here.

I have already calculated the following components, which are hopefully correct:

$$\Delta x = \frac 1n, \quad x_i = 1+i\Delta x, \quad f(x_i) = 5+6\left(1+\frac 1n\right)^2$$

Using this I was able to get to the following point: (1)\begin{equation} \begin{split} \frac 1n \sum_{i=0}^n f(x_i)&=\frac 1n \sum_{i=0}^n 5+6(1+\frac 1n)^2 \\ &= \frac 1n (\sum_{i=0}^n 11 + \sum_{i=0}^n \frac {6i^2}{n^2} + \sum_{i=0}^n \frac {12i}{n}) \end{split} \end{equation}

However my final answer after a few more steps was shown as incorrect. $$12n + \frac {6}{n^3}\frac n6 (2n+1)(n-1) + \frac{6(n+1)}{n}$$

I have obviously left out a few steps of my working, but I didn't want to bloat the question too much. Assuming my summations are correct at the end of (1), I then swapped things around to get the second and third as summations of $i$ and $i^2$, and then used the relevant rules to get my answer. I can edit to add all of my working if needed.

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1 Answer

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You have several mistakes in your first row of expressions. For the general expressions in my answer below, let the integral be $\int_a^b f(x)\,dx$.

Your $\Delta x$ is correct, being

$$\Delta x=\frac{b-a}n=\frac{1-0}n=\frac 1n$$

However, your $x_i$ is wrong and should be

$$x_i=a+i\Delta x=0+i\Delta x=i\Delta x$$

Your $f(x_i)$ is also wrong and should be

$$f(x_i)=5+6\left(\frac 1n\right)^2$$

Your formulas for $x_i$ and $f(x_i)$ seem to assume that the limits of integration are from $1$ to $2$, but the one you wrote is from $0$ to $1$.

Also, the limits on your summations are wrong. You have them from $0$ to $n$, but they should either be from $0$ to $n-1$ (as in the answer you linked to, which method is called the "left-side rule" or "left rectangular approximation method" or LRAM) or from $1$ to $n$ (which is the "right-side rule" or "right rectangular approximation method" or RRAM). So you want

$$\frac 1n\sum_{i=0}^{n-1} f(x_i) \qquad\text{or}\qquad \frac 1n\sum_{i=1}^{n} f(x_i)$$

Since your function is increasing, the lower sum $L_n$ will be the first summation and the upper sum $U_n$ will be the second.

Correct those errors and try again.

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