Question

I need help with integrals. I don't understand the basic rules. Therefore, I cant evaluate. Help would be much appreciated.

Answer 1

What do you not understand??

Answer 2

I was absent when my teacher introduced integrals. I don't know how to evaluate.

Answer 3

For example: \[\int\limits_{1/2}^{3} (2-\frac{ 1 }{ x }) dx\]

Answer 4

the integral is the anti derivative..so if you have y=f(t) the integral of f'(t)=f(t)..

Answer 5

for that problem you can integrate is a difference..so this case integral of 2dx - integral 1/xdx

Answer 6

I thought I was supposed to find the antiderivative of inside the parentheses and evaluate that or something..sigh..idk.

Answer 7

you need to look up the rules and see how they are defined using the textbook..if you understand derivatives you shouldnt have a problem with these introductory problems

Answer 8

I have my textbook. I don't understand. That is why I am asking for help. Yet you're telling me to consult my textbook. Gee, thanks.

Answer 9

did u learn F(b) - F(a) yet?
if u did, just do antiderivative of the stuff inside the parenthesis
then do F(b) - F(a)
for example,
\[\int\limits_{a}^{b} f(x)dx\]
say F(x) is antiderivative of f(x)
then it'd be F(b) - F(a)

Answer 10

Yes, I did learn F(b)- F(a).

Answer 11

then just find antiderivative of inside and do F(b) - F(a)

Answer 12

is the antiderivative 2x-lnx ?

Answer 13

sorry i haven't gotten to ln's yet. but i can tell u that the 2x part is correct.

Answer 14

Um, ok. So if you took the antiderivative of 2- (1/x) what would you get?

Answer 15

Remember that \(\frac{d}{dx} [2x+c] = 2\) and that \(\frac{d}{dx}\ln(x) = 1/x\)

Answer 16

Right, so I got 2x-ln(x)

Answer 17

So \(\frac{d}{dx} 2x - \ln (x) + C = 2-1/x\)

Answer 18

Yes, exactly

Answer 19

Ok, but now what do I do?

Answer 20

\[
b = 2 \\
a = 1/2 \\
F(x) = 2x - \ln(x) +C \\
\implies \\
F(b) - F(a) = [2(2) - \ln(2) +C] - [2(1/2) - \ln(1/2) +C]

\]

Answer 21

b=3

Answer 22

So, [2(3)-ln(3)+C] - [2(1/2)-ln(1/2)+C] ?

Answer 23

technically, you don't really need to worry about the C's because they'll cancel out.

Answer 24

and yes. if all your other steps are right.

Answer 25

Can you help me simplify? I think im doing it wrong

Answer 26

what did u get?

Answer 27

umm..[6-ln(3)] - [1-ln(1/2)]

Answer 28

yeah that's what i got too.

Answer 29

Is there a way to simplify further?

Answer 30

http://www.wolframalpha.com/input/?i=integrate+%282-1%2Fx%29+from+1%2F2+to+3
the log should be a ln

Answer 31

Ohh, ok, thanks.