- #1

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So, I know this is a pretty simple problem, but I seem stuck on it nevertheless.

Here's the question

Calculate the upper and lower sums , on a regular partition of the intervals, for the following integrals

\begin{align*}

\int_{1}^{3}(1-7x)dx

\end{align*}

Please correct me if I'm doing this wrong. Here's what I did

\begin{align*}\Delta x = \frac{3-1}{n} = \frac{2}{n}\end{align*}

\begin{align*}x_{i}=1+\Delta x*i = 1 + \frac{2}{n}i\end{align*}

\begin{align*}f(x_{i}) = (1-7x_{i})) = (1-7(1+\frac{2}{n}i))\end{align*}

\begin{align*}\frac{2}{n}\sum_{i=1}^{n}f(x_{i}) = \frac{2}{n}\sum_{i=1}^{n}(1-7(1+\frac{2}{n}i))\end{align*}

Considering the question asked for it on a regular partition of intervals, I assumed that meant 3, so I set n=3. Doing this, I get the answer -92/3, however apparently this isn't the correct answer for the lower sum. If I run n up to approach infinity, it goes to -26, which seems right, but that isn't accepted either.

Have I done something terribly wrong? Thanks