The Student Room Group
Reply 1
What...?

Do you want to evaluate the integral because you are effectively there...
If you have [stuff]ba\displaystyle \left[ \text{stuff} \right]^a_b then you do stuff(a)Β -Β stuff(b)\text{stuff(a) - stuff(b)}.

So it's βˆ’46Γ—26βˆ’(βˆ’46Γ—16)\dfrac{-4}{6} \times 2^6 - (\dfrac{-4}{6} \times 1^6).
Reply 3
Jimny
What...?

Do you want to evaluate the integral because you are effectively there...

complete the following integrations, it says

i'm a maths noob :p: :o:
robinson999
complete the following integrations, it says

i'm a maths noob :p: :o:


Ah. Well, that general rule is:

∫xn=xn+1n+1\displaystyle \int x^n = \frac{x^{n+1}}{n+1} (for n not equal to 1).

In your case n = 5, so insert this into the formula, remembering the multiplier of 4 and then use my post above.

Maths repulses me. :frown:
Reply 5
i think i get it, i don't think my lecture notes help at all
can't see where the other numbers came from in the example they use

βˆ’46Γ—26βˆ’(βˆ’46Γ—16)\dfrac{-4}{6} \times 2^6 - (\dfrac{-4}{6} \times 1^6)

would they just come out as -42?
or 64
i'll take -42 as the post below says :p:
Reply 6
yep
Glutamic Acid
Ah. Well, that general rule is:

∫xn=xn+1n+1\displaystyle \int x^n = \frac{x^{n+1}}{n+1} (for n not equal to 1).

In your case n = 5, so insert this into the formula, remembering the multiplier of 4 and then use my post above.

Maths repulses me. :frown:

GA do you know how to derive it? I know it should involve Riemann sum notation.
Simplicity
GA do you know how to derive it? I know it should involve Riemann sum notation.


I can do it using inequalities (there was an old question on find x^5; it could be easily generalized for x^n), not sure about Riemann sum notation, though. [Actually, the aforementioned question is a definite integral between 0 and 1, but I think that's fixable.]
Glutamic Acid
(for n not equal to 1)

*stabs*

n not equal to -1.
generalebriety
*stabs*

n not equal to -1.


Oh, oops. :o:

(+1)

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