Integration of logs and exponentials questions (1 Viewer)

Kittikhun

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a). Use the trapezoidal rule with 5 function values to approximate

\int_{0}^{2}ln(2x+1)dx

Answer to three decimal places.

b). i) Find \int_{0}^{ln5}\left ( \frac{e^{y}-1}{2} \right )dy

ii). Hence, find the exact value of


\int_{0}^{2}ln(2x+1)dx


I have no idea how to use the Latex thing so please forgive me and insert the command or whatever into the equation thing please.

Thanks.
 
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sazlik

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For (a):



Will give the others a shot in a sec. :)
 

sazlik

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For (b), part i. :



The second part of (b) is proving a bit of a challenge for me at this hour, but I hope that's somewhat helpful. I'm hoping I haven't made too many mistakes; please feel free to correct me if you spot any.

Thanks for the practice! :)
 
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Kittikhun

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Thanks! :)

Goddamnit, the first one is piss easy. What the hell was I thinking trying to integrate it using the special result of xlnx-x+C for I= Inx? I'm stuffed.

I don't know about the second one since I have the past paper but not the answer. It's stupid, I know. It's just that keyword hence that is pissing me off.
 

sazlik

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Goddamnit, the first one is piss easy. What the hell was I thinking trying to integrate it using the special result of xlnx-x+C for I= Inx? I'm stuffed.
I think that's what you would do in the second part of (b). No wonder I couldn't figure it out—we haven't covered it yet, haha! :) It looks like it should be relatively straightforward, though. I got the wrong answer, but I forgot the function of a function rule or something and it should work otherwise. Still, I don't understand the 'hence' thing, either. Odd.

I don't know about the second one since I have the past paper but not the answer.
I'm pretty sure my answer was right—I think I tested it in an online integral calculator. :) My new favourite discovery!
 

kcqn93

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wow that's pretty impressive for year11.

i didn't learn integration til term 4 last year, not to mention logs and exp.... :S
 

Rezen

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For the last part, notice that

also, when y=ln5, x=2. when y=0, x=0. Since the integral is the area underneath the function, (when its with respect to y, its the area the function makes with the y-axis)

 

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