how to integrate this? (2 Viewers)

[jannny]

y` = 1 / ( 20t - 5t^2)

thanks =]

 

[williamc]

y` = 1 / ( 20t - 5t^2)

y'= 1 / (20t-5t^2)


y'= (20t-5t^2) ^-1


y= (20t-5t^2) ^-1 / -1 * 20.10t (function of a function rule)


y= (20t-5t^2)^-1 / -1 * (200t)


y= (20t-5t^2)^-1 / -200t


y= 1 / (20t-5t^2) / -200t


y= 1/ (20t-5t^2) * 1/-200t


y= (20t-5t^2) ^-1 / -200t ^-1 + C


ermm i think i fludged somewhere

 

[Forbidden.]

I thought you integrated to a natural logarithm (log_e f(x))

 

[williamc]

I thought you integrated to a natural logarithm (log_e f(x))

Oh shit i forgot.. thats what you do.

Edit: Wait maybe not

 

[jannny]

mmm seems wrong, there's no function of a function rule for integration right?

 

[williamc]

mmm seems wrong, there's no function of a function rule for integration right?

Yes there is

 

[jannny]

so wat would the working out be?

 

[williamc]

Actually wait.. i think when you do function of a function rule.. you have to add one to the orginal power.. which would make it equal to the ^ of 0. Which there is an error.

Edit: i'll get my textbook for the formula.

 

[Forbidden.]

mmm seems wrong, there's no function of a function rule for integration right?

Yes there is

ya ...

∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ / a(n+1) + C

... I think ...

Actually wait.. i think when you do function of a function rule.. you have to add one to the orginal power.. which would make it equal to the ^ of 0. Which there is an error.

Edit: i'll get my textbook for the formula.

Above ?

Also ... ∫ 1/x dx = ∫ x^-1 dx MUST equal ln x + C, you can't have x⁰ + C

 

[nichhhole]

ya ...

∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ / a(n+1) + C

... I think ...



Above ?

Also ... ∫ 1/x dx = ∫ x^-1 dx MUST equal ln x + C, you cqn't have x⁰

but doesnt the original function have to be linear?

 

[williamc]

Maybe that function can't be integrated. Or have to use another rule/formula.

 

[williamc]

but doesnt the original function have to be linear?

We only use linear equations in 2 unit i thought.

 

[nichhhole]

thats what i thought..
but the orig eq was

1/[20t-5t^2]

 

[Forbidden.]

but doesnt the original function have to be linear?

?

 

[jannny]

How bout if I do like this:

∫ 1/(20t-5t^2) dt

... Which could be equal to... (correct me if im wrong)

∫ (1/20t) - 5t^-2 dt

( 1/20 ) ∫ (20/20t) - ∫(5t^-2)

= 1/20 * ln (20t) + 5 / t + C

not sure =/

 

[Forbidden.]

We only use linear equations in 2 unit i thought.

Fractions too as long as they can be integrated to a natural logarithm,
like ∫ 5/3x dx = (5/3) ln (3x) + C, not something crazy like the OP posted.

 

[williamc]

∫ (1/20t) - 5t^-2 dt

( 1/20 ) ∫ (20/20t) - ∫(5t^-2

= 1/20 * ln (1/20t) + 5 / t + C

not sure =/

Errmm cough *wtf * cough

 

[Forbidden.]

Who's going to try integration by substitution ?

 

[williamc]

Fractions too as long as they can be integrated to a natural logarithm,
like ∫ 5/3x dx = (5/3) ln (3x) + C, not something crazy like the OP posted.

Yer. I'm pretty sure that question isn't a 2unit question, more like a 'harder 2 unit question.'

 

[jannny]

huh? wtf is inverse tan .

Can some1 post a completed workout?