Author Topic: 4U Maths Question Thread (Read 665378 times) Tweet Share

VydekiE · « **Reply #1080 on:** April 13, 2017, 05:33:25 pm »

Hi, it would be great if I could get some help on this integration question
1. Integrate 1/5+4x+x² dx
Thank you!!

RuiAce · « **Reply #1081 on:** April 13, 2017, 06:44:12 pm »

Quote from: VydekiE on April 13, 2017, 05:33:25 pm

Hi, it would be great if I could get some help on this integration question
1. Integrate 1/5+4x+x² dx
Thank you!!

Is there a bracket missing around the denominator?

(If yes, hint: complete the square)

VydekiE · « **Reply #1082 on:** April 13, 2017, 07:55:28 pm »

Quote from: RuiAce on April 13, 2017, 06:44:12 pm

Is there a bracket missing around the denominator?

(If yes, hint: complete the square)

Nope there's no brackets missing

RuiAce · « **Reply #1083 on:** April 13, 2017, 08:01:45 pm »

VydekiE · « **Reply #1084 on:** April 14, 2017, 04:13:57 pm »

Quote from: RuiAce on April 13, 2017, 08:01:45 pm

$\text{If no brackets are missing then this is a 2U integral.}\\ \int \frac15 + 4x + x^2\, dx = \frac{x}{5} + 2x^2 + \frac{x^3}{3}+C$

Didn't think of that. Thank you!!

hanaacdr · « **Reply #1085 on:** April 15, 2017, 07:17:52 am »

Hi
could i please get some help on part b please
thanks

jakesilove · « **Reply #1086 on:** April 15, 2017, 02:48:21 pm »

Quote from: hanaacdr on April 15, 2017, 07:17:52 am

Hi
could i please get some help on part b please
thanks

We know that the net force, on the way down, will be

$ma=mg-mkv$
$a=g-kv$
$\frac{dv}{dt}=g-kv$
$\frac{dv}{g-kv}=dt$

$\int_0^w\frac{dv}{g-kv}=\int_0^{T'}dt$

$-\frac{1}{k}ln(\frac{g-kw}{g})=T'$

$-T'k=ln(\frac{g-kw}{g})$

Also,

$v\frac{dv}{dx}=g-kv$

$\frac{vdv}{g-kv}=dx$

$-\int_0^w\frac{g}{k}\frac{1}{kv-g}+\frac{1}{k}dv=\int_H^0dx$

$\frac{g}{k}ln(\frac{kw-g}{-g})+\frac{w}{k}=H$

Subbing in what we found the first time,

$\frac{g}{k}(-T'k)+\frac{w}{k}=H$

And that should get you to the answer. Possible I messed up a sign/factor somewhere, but that's the general set-up

hanaacdr · « **Reply #1087 on:** April 15, 2017, 08:28:39 pm »

thank you!

Quote from: jakesilove on April 15, 2017, 02:48:21 pm

We know that the net force, on the way down, will be

$ma=mg-mkv$
$a=g-kv$
$\frac{dv}{dt}=g-kv$
$\frac{dv}{g-kv}=dt$

$\int_0^w\frac{dv}{g-kv}=\int_0^{T'}dt$

$-\frac{1}{k}ln(\frac{g-kw}{g})=T'$

$-T'k=ln(\frac{g-kw}{g})$

Also,

$v\frac{dv}{dx}=g-kv$

$\frac{vdv}{g-kv}=dx$

$-\int_0^w\frac{g}{k}\frac{1}{kv-g}+\frac{1}{k}dv=\int_H^0dx$

$\frac{g}{k}ln(\frac{kw-g}{-g})+\frac{w}{k}=H$

Subbing in what we found the first time,

$\frac{g}{k}(-T'k)+\frac{w}{k}=H$

And that should get you to the answer. Possible I messed up a sign/factor somewhere, but that's the general set-up

Rathin · « **Reply #1088 on:** April 16, 2017, 12:59:58 pm »

Can someone check if my method is correct?

Q) I=∫=x^3*e^x^2dx

A)

u=x^3
u'=3x^2
v'=e^x^2
v=2xe^x^2

Therefore;
I=2x^4*e^x^2 - 6∫x^3*e^x^2dx
7I= 2x^4*e^x^2 +C
I= (2x^4*e^x^2)/7 +C

RuiAce · « **Reply #1089 on:** April 16, 2017, 01:13:33 pm »

Quote from: Rathin on April 16, 2017, 12:59:58 pm

v'=e^x^2
v=2xe^x^2

Look here. You differentiated your v', instead of integrated it.
$\text{The most optimal choice for an integration by parts would be}\\ u=x^2 \implies du = 2x\, dx\\ dv = xe^{x^2} \, dx \implies v = \frac12 e^{x^2}$

Rathin · « **Reply #1090 on:** April 16, 2017, 01:19:58 pm »

Quote from: RuiAce on April 16, 2017, 01:13:33 pm

Look here. You differentiated your v', instead of integrated it.
$\text{The most optimal choice for an integration by parts would be}\\ u=x^2 \implies du = 2x\, dx\\ dv = xe^{x^2} \, dx \implies v = \frac12 e^{x^2}$

Omg these silly mistakes..

beau77bro · « **Reply #1091 on:** April 18, 2017, 01:22:11 pm »

Q15 bii, I don't really get it, how do we do simultaneous, i feel I missed something simple

RuiAce · « **Reply #1092 on:** April 18, 2017, 02:09:30 pm »

RuiAce · « **Reply #1093 on:** April 18, 2017, 02:12:16 pm »

If you meant a) ii it's actually just rearranging.
$b^2=a^2-75\text{ so substituting into the other one}$
$\frac{64}{a^2}+\frac{9}{a^2-75}=1\\ 64(a^2-75) + 9a^2 = a^2(a^2-75)\\ \text{This is just a quadratic in }a^2.\text{ You may want to consider the quadratic formula.}$

Rathin · « **Reply #1094 on:** April 18, 2017, 10:06:50 pm »

Help with this please
∫sin(x)sin(3x)dx

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