HELP! Can someone integrate these? (1 Viewer)

[tooheyz]

hey guys.. most of u should know from the other threads that i totally hate logs with a passion! LOL

well here are a few questions which i cant do... so if any can help... please do!

thanks in advance

find deriv. of (e^x2) <--- thats x^2

and hence evaluate (the integral from 1 and 0) S of 1-0 xe^x2 <-- thats x^2

 

[kevinexcs]

(e^x2) <--- thats x^2

e^u=u'e^u

2xe^x2<--- thats x^2
not sure about that. and for the others...can't think right now ill try to do it later..

 

[tooheyz]

yeah thats right. i got the first part right as well
after like an hour of analysing it ..

well i've been trying to get the second part right.. i cant get it!

when u intergrate xe^x2 between 1 and 0, it should be 1/2 (e-1) and should be approx 0.86.

yeah thats the answer and i cant get it!

 

[tooheyz]

hahah yeah i got it! FINALLY!

yeah .. i use the HENCE part.. at the beginning i didnt, i tried to integrate it as it was LOL

yeah theres another question which i cant do...

umm... can someone help me? [sorry to nagg]

(all to base e)

log 2 = 0.69
log 5 = 1.61

find log (5/2e)

thanks in advance again

 

[kevinexcs]

Log base e same as In easy for me to look at

well... In(5/2e) using log laws...
In5 - (In2 +Ine)
1.62-(.69+1)
not sure..

 

[tooheyz]

hahah thats right! yeah i was doing the same thing but i thought iwas doing the wrong thing

ah well... thanks mate! your a legend!

 

[Saintly Devil]

Another question regarding integration.....

Which one is the answer to the indefinite integral of d(x^3)/dx (i.e. the integration of the differentiation of x cubed):

1) x^3 +C
2) x^3


I thought it was 2), since you already know the initial 'y' value, and so by integrating it's derivative, there is really no unknown.
But my teacher said it was x^3 + C. Can anyone explain why this is/isn't the case?

Thanks

 

[tooheyz]

whenever you integrate something.. there is ALWAYS a contant after the integral.. ALWAYS

so just add the constant 'K' or 'C'

but i dont know.. i dont exactly understand your questions

 

[Huy]

Re: Another question regarding integration.....

Originally posted by Saintly Devil
Which one is the answer to the indefinite integral of d(x^3)/dx (i.e. the integration of the differentiation of x cubed):

1) x^3 +C
2) x^3


I thought it was 2), since you already know the initial 'y' value, and so by integrating it's derivative, there is really no unknown.
But my teacher said it was x^3 + C. Can anyone explain why this is/isn't the case?

Thanks

tooheyz, that was pretty confusing
but he meant the result when x^3 is differentiated, integrated

or the differential of x^3, integrated.
either way, differentiate first, and integrate
(what's the point! differentiating and reversing the process via integration... -1 to the power, +1 to the power ... amazing )



but anyway, the answer is 1)

because indefinite integrals have no 'restrictions' per se (terminals, end points) so you need the constant +C, +k as tooheyz stated.

d/dx x^3 = 3x^2 (simple enough)

integrate 3x^2 and you get 3x^3 on 3 (add 1 to the power, divide by the new power), and simplifying

3x^3
------- +C
3

= x^3 + C is your answer #

answer 2) would apply if you were given end points e.g. x = 1, x = 5 or whatever and you'd sub in the values, from a to b and
subtract as you would normally

 

[Twintip]

Originally posted by ToOhEyZ
whenever you integrate something.. there is ALWAYS a contant after the integral.. ALWAYS

so just add the constant 'K' or 'C'

No no no!!! ONLY if they are indefinite integrals. If they are definite intergrals you DO NOT put "+c" or whatever on the end.

 

[Minai]

jus of side interest...in the HSC, the examiners do not take marks off if u omit the "+C" at the end of indefinite integrals...but in school based assessments, ur teachers most likely will

 

[ND]

Originally posted by MinAi
jus of side interest...in the HSC, the examiners do not take marks off if u omit the "+C" at the end of indefinite integrals...but in school based assessments, ur teachers most likely will

Nice, i seem to lose quite a few marks because of that. Although i think (hope) i've learnt my lessons by now.

 

[tooheyz]

Originally posted by Twintip
No no no!!! ONLY if they are indefinite integrals. If they are definite intergrals you DO NOT put "+c" or whatever on the end.

LOL no WONDER i've been getting them wrong!
LOL thanks for clearing that up for me man!
LOL i've been adding a constant to all my answers... oh dear

 

[Twintip]

Lol I've forgotten the +c on some indefinite integrals in a few of my 3u assessments and lost a mark. MinAi they could theoretically dedicate a mark to the "+c" in an HSC exam question, though, couldn't they?

 

[Minai]

Originally posted by Twintip
Lol I've forgotten the +c on some indefinite integrals in a few of my 3u assessments and lost a mark. MinAi they could theoretically dedicate a mark to the "+c" in an HSC exam question, though, couldn't they?

no, they dont
simple integration questions requiring "+c" are usually worth only 2 marks
1 mark for the integration
1 mark for the answer

 

[Twintip]

Ah k.

 

[KeypadSDM]

You always add a c -- even to definite integrals...
watch:

/1
|f'(x)dx
/0

=[f(x) + c]1,0
=(f(1) + c) - (f(0) + c)
=f(1) - f(0)

So there

 

[tooheyz]

hey guys i got some more problems with integrating!

well theres 2 questions that r driving me up the wall

1. differentiate with respect to x. y = In x / (x - 2)



my answer was x - 2 - xInx / (x - 2 )*2


but the answer is x - 2 - xInx / 2 (x - 2 )*2


and another intergrating problem..... find intergral S 3*x dx.

thats 3 to the power of x....

thanks in advnace guys!

 

[McLake]

1. differentiate with respect to x. y = In x / (x - 2)

y = (ln x) / (x-2)

u = ln x
du = 1/x
v = x - 2
dv = 1

(vu' - uv') /v^2
= ((x-2/x) - ln x) / (x-2)^2
= (x-2-xlnx)/x(x-2)^2


2. find intergral S 3^x dx

let y = 3^x
log3y = x
use chang of base
log3y = ln y/ ln3
lny = xln3
y = e^(xln3)
dy/dx = ln3*e^(xln3)

 

[tooheyz]

ah thanks mate! your a legend