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May 22, 2025, 06:10:48 pm

Author Topic: Bazza's 3/4 Question Thread  (Read 33562 times)  Share 

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WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #60 on: June 24, 2012, 05:11:17 pm »
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How do i integrate 1/(1+sinx)

Its driving me insane



Ugh finally got it it D:

How can i  see what to integrate quickly ? Like what i should try and convert to?
« Last Edit: June 24, 2012, 05:18:45 pm by Bazza16 »

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Re: Bazza's 3/4 Question Thread
« Reply #61 on: June 24, 2012, 05:19:15 pm »
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How do i integrate 1/(1+sinx)

Its driving me insane

dw about it, you'll never get anything that complicated in any SAC/exam without prompting on what to do next.

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Re: Bazza's 3/4 Question Thread
« Reply #62 on: June 24, 2012, 06:04:43 pm »
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How did you do it? I'm pretty sure the solutions were wrong...
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WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #63 on: June 24, 2012, 06:17:01 pm »
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Yeah solutions were wrong, mr g thought that it was cosx

Uh "rationalise it" then split into 2 integrals then integrate

WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #64 on: June 25, 2012, 04:35:57 pm »
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what is the area bound by y=ln2x , axes and y=2?

i did e^y = 2x


and then integrated with respect to y 1/2S(2,0 terminals)e^y dy is what i put
is that correct?
(my teachers answers have something random)

b^3

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Re: Bazza's 3/4 Question Thread
« Reply #65 on: June 25, 2012, 04:42:28 pm »
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what is the area bound by y=ln2x , axes and y=2?

i did e^y = 2x


and then integrated with respect to y 1/2S(2,0 terminals)e^y dy is what i put
is that correct?
(my teachers answers have something random)
Thats they way I'd look at it, unless I've missed something. Ended up with square units.

What did your teacher do?
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WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #66 on: June 25, 2012, 04:49:01 pm »
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full question is
The area bounded by the curve y = ln2x , the coordinate axes, and the line y =2 is given by the integral :

he put C

S((e^2)/2, 1 terminals) ln2x dx

he put worked solutions up which make some reference to the "inverse" having the same amount under the x axes but i'm not sure why he did that or whether i'm missing somethign

Phy124

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Re: Bazza's 3/4 Question Thread
« Reply #67 on: June 25, 2012, 04:53:29 pm »
+1
I did;



Where the comes from



(same answer as b^3)

So yeah you guys are right (AFAIK)

edit: changed x to 2x
« Last Edit: June 25, 2012, 04:59:40 pm by The AN Dunce »
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Re: Bazza's 3/4 Question Thread
« Reply #68 on: June 25, 2012, 05:57:23 pm »
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They're both valid, except at spesh level, the first way can be done by hand, the second one cannot, needs a CAS calc (or by integration by recongition in methods (which is backwards integration by parts, but you don't need to know that now...... forget I said the thing in the second level brackets)).

So if its exam 1, then you would need to do the first one.
« Last Edit: June 25, 2012, 05:59:40 pm by b^3 »
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WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #69 on: June 25, 2012, 07:58:00 pm »
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it was exam 2, both were options on the SAC

i think the infallible Dr. G. has made 2 mistakes in 3 of the past SACs combined

one was having 2 correct options, another was asking to prove something incorrect :O

does anyone have a good 2 (double sided) page calculus cheat sheet by any chance :D

WhoTookMyUsername

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Re: Bazza's 3/4 Question Thread
« Reply #70 on: June 25, 2012, 08:15:39 pm »
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it was exam 2, both were options on the SAC

i think the infallible Dr. G. has made 2 mistakes in 3 of the past SACs combined

one was having 2 correct options, another was asking to prove something incorrect :O

does anyone have a good 2 (double sided) page calculus cheat sheet by any chance :D

Also

When it says evaluate the integral, do you need to account for when the equation crosses the x axis?
(im pretty sure not, but i almost made this mistake today, want to be sure)

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Re: Bazza's 3/4 Question Thread
« Reply #71 on: June 25, 2012, 08:19:37 pm »
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When it says evaluate the integral, you just do it without worrying about the graph interesting the x-axis. But if it asks specifically for the area, then you do account for it :)
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Re: Bazza's 3/4 Question Thread
« Reply #72 on: June 25, 2012, 09:14:51 pm »
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When it says evaluate the integral, you just do it without worrying about the graph interesting the x-axis. But if it asks specifically for the area, then you do account for it :)

yeah, don't equate integration and area calculation. integration is but a technique that facilitates the calculation of areas under curves. so when asked to evaluate a definite integral, don't even bother looking at graphs or adding units^2. just solve it as per usual.
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Re: Bazza's 3/4 Question Thread
« Reply #73 on: June 25, 2012, 09:44:56 pm »
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Why does the integral of a function give the area under the graph? Is there some sort of proof or explanation?
Help would be appreciated :)
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