Question

\[\int\limits_{0}^{2}(2x+1)dx\] Evaluate the definite integral

Answer 1

OK. Find the antiderivative, then plug in the values to evaluate. What part of this is giving you trouble?

Answer 2

really? that's not what the book is telling me to do.

Answer 3

What is the book telling you to do?

Answer 4

Use the definition of definite integral to evaluate the following definite integrals. Use right Riemann Sums and the sigma notation theories

Answer 5

Ouch. Still in the part where you can't use the shortcuts. OK. Then you end up having to do it the long way or they do not give you credit for the work.

Answer 6

God dammit there's a short way? Those B astards. I'm not even turning this in it's just for practice. Why would they do that?

Answer 7

So you learn to appreciate the shortcuts more, and so you know how to use the sums for the times when you don't have a shortcut.

Answer 8

Well do you think you could go through it with me this one time. I know it might take a while, so it's cool if you don't want to.

Answer 9

I've tried to work it out a couple times but i keep getting it wrong

Answer 10

OK. So first you divide it into n equal parts. It is from 0 to 2, so you have:\[\Delta x=\frac{2-0}{n}\]

Answer 11

you should get to this right?
\[\sum_{k+1}^{n}(2(\frac{ 2k }{ n })+1)\frac{ 2 }{ n }\]

Answer 12

which would be

Answer 13

\[\frac{ 2 }{ n }\sum_{k=1}^{n}(\frac{ 4k }{ n }+1)\]

Answer 14

and then from there you would have \[\frac{ 2 }{ n }(\frac{ 4 }{ n } \sum_{k=1}^{n}k+\sum_{k=1}^{n}1)\]

Answer 15

then \[\frac{ 2 }{n }[\frac{ 4 }{ n }\frac{ n(n+1) }{ 2 }+n]\]

Answer 16

then I guess it would be \[\frac{ 4n+4 }{ n }+n\] right?

Answer 17

What did you do when you distributed the 2/n into the entire thing? Your +n tells me that might be the issue.
\[\frac{ 8 }{ n^2 }\frac{ n(n+1) }{ 2 }+\frac{2n}{n} \]

Answer 18

The left part of that does reduce to what you have. But the right...

Answer 19

oh I see forgot to distribute to the n

Answer 20

yah. Now, want to be blown away by the simple form?

Answer 21

stupid mistakes. Seems like that is always my problem. Yes please I would love to see that

Answer 22

\[\int\limits_{0}^{2}(2x+1)dx \Rightarrow\\
|_0^2(x^2+x) \Rightarrow\\
2^2+2-[0^2+0] \Rightarrow\\
\int\limits_{0}^{2}(2x+1)dx = 6\]

Answer 23

wow! lol super easy huh?

Answer 24

Yah, and you just did all that work.... and it is like, the top minus the bottom. The +c goes away because top+C - [bottom+C] = top-bottom+C-C

Answer 25

Ah yes I see they discuss it in the very next section in my book

Answer 26

Anyway thanks so much for your time.

Answer 27

Yah. Then you go on to other forms of integration and applications of it. Like how this suddenly tels you how to find volumes of strange shapes. Remember optimization? Now you can sudenly optimize an odd shape to find the best possible item to make that does what is needed. All sorts of practical applicatons.

Answer 28

Oh, and very soon you can do proofs of all the geometry volume formulas.

Answer 29

That's what my teacher says

Answer 30

Yah, math profs. LOVE proofs..... and if you go for a degree in Mathematics, that will be a huge chunk of it. For people going into other sciences, they can escape a lot of that part of math because they don't have to take all those classes.