Why is the summation in the first part from k=0 to n-1?

[VikramAhuja]

∫(4−7x)dx ====> Integral is from 1 to 8

Similar question but for some reason I can't get the answer after following all the steps

 

[MarkFL]

∫(4−7x)dx ====> Integral is from 1 to 8

Similar question but for some reason I can't get the answer after following all the steps

I have moved your post to its own thread so that the other thread does not become convoluted.

Using a similar argument in my first post of the other thread, we can apply some simple geometry to see that the result should be:

\(\displaystyle I=\int_1^8 4-7x\,dx=-\left(3\cdot7 + \frac{1}{2}7\cdot49\right)=-\frac{385}{2}\)

Now, in setting up the left-hand sum, we may begin with:

\(\displaystyle I_n=\frac{7}{n}\sum_{k=0}^{n-1}\left(4-7\left(1+k\left(\frac{7}{n}\right)\right)\right)\)

Are you with me so far?

 

[VikramAhuja]

Yes I've got to (7/n^2)((43n^2-49n)/2)

Apparently it is wrong

 

[MarkFL]

Okay, my next step would be:

\(\displaystyle I_n=-\frac{7}{n^2}\sum_{k=0}^{n-1}\left(3n+49k\right)\)

And next, I would write:

\(\displaystyle I_n=-\frac{7}{n^2}\left(3n\sum_{k=0}^{n-1}(1)+49\sum_{k=0}^{n-1}(k)\right)\)

Using summation formulas, there results:

\(\displaystyle I_n=-\frac{7}{n^2}\left(3n^2+49\frac{n(n-1)}{2}\right)\)

Now, factor out a \(\displaystyle \frac{n}{2}\), distribute and combine like terms. What do you find?

 

[VikramAhuja]

Nevermind I got it but now with lower sums?

 

[MarkFL]

Would the answer be (-7(55*n-49))/(2*n)

Yes, I also get:

\(\displaystyle I_n=-\frac{7(55n-49)}{2n}\)

How'd you get this? (-7/n^2) ?

We initially had \(\displaystyle -\frac{7}{n}\) in front of the sum as a factor, and then I factored out \(\displaystyle \frac{1}{n}\) and then after combining terms, the summand became $3n+49k$.

 

[VikramAhuja]

Sorry I edited my post

 

[MarkFL]

Nevermind I got it but now with lower sums?

Can you show what you think the sum should initially be?

 

[VikramAhuja]

No I'm rather confused with lower sums

 

[MarkFL]

The right-hand sum will simply be:

\(\displaystyle I_n=\frac{7}{n}\sum_{k=1}^{n}\left(4-7\left(1+k\left(\frac{7}{n}\right)\right)\right)\)

 

[VikramAhuja]

Thanks a lot mate.

 

[VikramAhuja]

One last question. Why is the summition in the first part sum_{n-1}^{k=0}

 

[MarkFL]

One last question. Why is the summition in the first part sum_{n-1}^{k=0}

In the left-hand sum, the height of each rectangle is taken to be the integrand's value on the left side of each regular partition, so the index of summation needs to go from $k=0$ to $k=n-1$.