Solve Upper & Lower Sums: Step-by-Step Guide

[reefster98]

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.

∫(3−5x)dx ======> integral is from 1 to 7

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.

Someone please help me.

My Lower sum working out:

x_i= 1 + iΔx = 1 + 6i/n

To calculate the lower sum, I used the rule Δx\sum_{i=1}^{n}

f(x_i) = 3 - 5(1 + 6i/n) = (-30i-2n)/n

substituting it into the sum rule stated above, my answer became 6/n(-17n - 15) = -42 -90/n

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

Please help me solve this.

 

[MarkFL]

Using a little geometry, and the fact that all of the area is below the $x$-axis, we know what we need to find is:

\(\displaystyle I=\int_1^73-5x\,dx=-\left(6\cdot2+\frac{1}{2}\cdot6\cdot30\right)=-102\)

Okay, we now have a value that we can use as a check.

For the left-hand sum, I would state:

\(\displaystyle I_n=\frac{6}{n}\sum_{k=0}^{n-1}\left(3-5\left(1+k\left(\frac{6}{n}\right)\right)\right)\)

\(\displaystyle I_n=-\frac{12}{n^2}\sum_{k=0}^{n-1}\left(n+15k\right)\)

\(\displaystyle I_n=-\frac{12}{n^2}\left(\sum_{k=0}^{n-1}\left(n\right)+15\sum_{k=0}^{n-1}\left(k\right)\right)\)

\(\displaystyle I_n=-\frac{12}{n^2}\left(n^2+15\cdot\frac{n(n-1)}{2}\right)\)

\(\displaystyle I_n=-\frac{6}{n^2}\left(2n^2+15n(n-1)\right)\)

\(\displaystyle I_n=-\frac{6}{n^2}\left(17n^2-15n\right)\)

\(\displaystyle I_n=-\frac{6(17n-15)}{n}\)

Hence:

\(\displaystyle I=\lim_{n\to\infty}I_n=-6\lim_{n\to\infty}\left(17-\frac{15}{n}\right)=-6(17)=-102\)

Can you now do the right-hand sum in a like manner?

 

[reefster98]

Thank you MarkFL for answering! Your answer and mine were both correct. I just had a syntax typing error (Wondering) haha