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Calculus

ughh..

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Christmas 2013
I'm working on some problems and i'm drawing a blank. I left my class notes somewhere and book examples are not doing it for me. Thanks in advance!

Mind starting me in the right direction with just 13?
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And if you want to check if you're right
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Number 13 you're going to want to take the anti derivative of the entire function. Which is the reverse of the power rule. Therefore giving you (1x^2)/4+3x then to evaluate this plug in your upper limit of integration giving you the result 16 if my mental math is not mistaken. Now you must plug in the lower limit of integration which means you must flip the integration limit to make -2 on top as your upper limit of integration. This flips the sign of the overall result of the integration. Therefore using the same anti derivative as before we would get -[(-2)^2 / 4 + 3(-2) which yields -5. Combining the two or taking the upper integration limit - the lower integration limit yields 16-(-5) = 21. Hope this helps.
 
Christmas 2013
Number 13 you're going to want to take the anti derivative of the entire function. Which is the reverse of the power rule. Therefore giving you (1x^2)/4+3x then to evaluate this plug in your upper limit of integration giving you the result 16 if my mental math is not mistaken. Now you must plug in the lower limit of integration which means you must flip the integration limit to make -2 on top as your upper limit of integration. This flips the sign of the overall result of the integration. Therefore using the same anti derivative as before we would get -[(-2)^2 / 4 + 3(-2) which yields -5. Combining the two or taking the upper integration limit - the lower integration limit yields 16-(-5) = 21. Hope this helps.

So like this if I were to write it out?
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Ah thank you! Now that I look at it it doesn't look difficult at all
 
Just a tip, and especially if you're on harder functions that have multiple terms like, say, [int( 3/ln(x)^2 + 1/sqrt(4-x^2))] it helps a lot to split the two up into separate integrals.

In number 3's case, you'd be looking at [int(x/2) + int(3)] with (-2,4) being the limits for both functions, which is easier to deal with than [int(x/2+3)]. Feel free to PM me if you need any more calc help :)
 
Christmas 2013
Just a tip, and especially if you're on harder functions that have multiple terms like, say, [int( 3/ln(x)^2 + 1/sqrt(4-x^2))] it helps a lot to split the two up into separate integrals.

In number 3's case, you'd be looking at [int(x/2) + int(3)] with (-2,4) being the limits for both functions, which is easier to deal with than [int(x/2+3)]. Feel free to PM me if you need any more calc help :)

Do you mean something like this? (Just added the graph to show what I was thinking. Could be done in my head)
jKt0bOLh.jpg
 
Do you mean something like this?

Yep, exactly like that :)

For addition and subtraction, you can use that method with any integral expressions.
 
sorry but nice hand writing... did you get to know how to work them out?
 
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