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Help with Intergration

ℓove

Hi.

Could someone explain how I can find k?
I expanded out the bracket and... do I intergrate by parts ? :s-smilie:

Thanks.

Reply 1

Notnek

Original post by ℓove

Hi.

Could someone explain how I can find k?
I expanded out the bracket and... do I intergrate by parts ? :s-smilie:

Thanks.

Since k is a constant, you have something of the form

ae^{bx} + ce^{dx}

Does that help? You don't need integration by parts - it's much simpler than that.

Reply 2

ℓove

I still don't understand -I'll just ask my teacher tomorrow.

Thank you anyway.

Reply 3

Notnek

Original post by ℓove

I still don't understand -I'll just ask my teacher tomorrow.

Thank you anyway.

Okay but if you still want to have a go today:

If you had

e^{2x}

then that integrates to give

\frac{1}{2}e^{2x} (+c)

So if you have

(4k-1)e^{2.5x}

then that integrates to give

\frac{1}{2.5}\times (4k-1)e^{2.5x}

And the second term can be integrated in a similar way.

Reply 4

ℓove

Okay it's coming back to me now. :smile:

So I got k=4.36 rounded to 3 s.f. but the answer says k=4 ...would you know if I went wrong somewhere?

Reply 5

NotNotBatman

Original post by ℓove

Okay it's coming back to me now. :smile:

So I got k=4.36 rounded to 3 s.f. but the answer says k=4 ...would you know if I went wrong somewhere?

You have subtracted a negative value, so you need to add 2/5 and 2k.

Reply 6

ℓove

Ooohh right!! :smile:

)))) I understand now. Thank you both!

(edited 6 years ago)

Reply 7

ℓove

Original post by r3035

....................so not a racist thread about migrants integrating into society ?

Hahahahah good one lol

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