Calculate the upper sums Un and lower sums Ln, on a regular partition of the intervals, for this integral:

sorry new to this site. Can someone please help me with this? I have tried for such a long time and have yielded no correct answers.

$$\int_1^7 (3−5x)dx$$

We have $n$ rectangles, so what I did first was found the change in $x$, which was $6/n$ which is the width of the rectangles. So Δx= $6/n$

I used summation to find the lower sum and upper sum but my answers were wrong.

My work:

$x_i = 1 + i\Delta x = 1 + 6\frac in$

To calculate the lower sum, I used Lower sum $= \Delta x\sum_{i=1}^n f(x_i)$

$$\begin{split} f(x_i) &= 3 - 5(1 + 6\frac in) \\ &= \frac{-30i-2n}{n} \end{split}$$ substituting it into the sum rule stated above, my answer became $$\frac 6n(-17n - 15) = -42 -\frac{90}{n}$$

This was wrong and I did almost the same for the upper sum too but that too is wrong.

• What were your answers? Why do you think they are wrong? – davidlowryduda Apr 25 '15 at 5:13
• To understand what you did wrong, you need to show us what you did. – user99914 Apr 25 '15 at 5:14
$$f(x_i) = 3 - 5 (1+ 6 \frac in).$$
$$\begin{split} \frac 6n \sum_{i=1}^n f(x_i) &= \frac 6n \sum_{i=1}^n \left( 3 - 5(1+ 6\frac in) \right)\\ &= \frac 6n \sum_{i=1}^n \left( -2 + 30\frac in \right)\\ &= \frac 6n \sum_{i=1}^n -2 + \frac 6n \sum_{i=1}^n 30\frac in \\ &= -12 + \frac{180}{n^2} \sum_{i=1}^n i \\ &= -12 + \frac{180}{n^2} \frac{n(n+1)}{2} \\ &= -12 + \frac{90(n+1)}{n}. \end{split}$$