Integration of logs and exponentials questions (1 Viewer)

[Kittikhun]

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.

 

[sazlik]

For (a):



Will give the others a shot in a sec.

 

[sazlik]

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!

 

[Kittikhun]

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]

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!

 

[hscishard]

I got 2.804 for the 2nd one.
e^in5 = 5.

Shit. - In5. Nvm you're right.

 

[kcqn93]

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]

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)