calculate ∫4x^3dx

increase power by one, divide by new power.

Also, to the second question, x^4 differentiates to 4x^3

$\displaystyle\int{ax^n}=\dfrac{ax^{n+1}}{n}+c$
For this case a=4, n=3

This.

This is the rule for integrating anything which is a multiple of a power of x.

Also, everyone so far has forgotten +C... (edit: this was true when I started typing, but Small123 has since proved me wrong)

$\displaystyle\int{ax^n}dx=\dfrac{ax^{n+1}}{n}+c$
For this case a=4, n=3

I though it was

$\displaystyle\int{ax^n}dx=\dfrac{ax^{n+1}}{n+1}+c$

??

What does x⁴ differentiate to?

Also, to the second question, x^4 differentiates to 4x^3

So many fails in one post , thanks !

Ehh.. apparently I should increase the power by one which would go to make 4^5 but then I divide by the new power? what does that mean/ which bit am i meant to divide? and what about the 3dx bit :s

∫4x^3dx

the DX bit simply means intergrate with respect to x.. this part will make more sense later on but for now leave it alone

youre meant to intergrate all the terms involving x in this expression.. so if you apply the rule to 4x^3, what do you think youll get? raise the power on the x term and then divide by the new power on the x term and see what happens

Ehh.. apparently I should increase the power by one which would go to make 4^5 but then I divide by the new power? what does that mean/ which bit am i meant to divide? and what about the 3dx bit :s

When integrating, the way how i remember is is that i add 1 to the power of the term, then divide it by the new power. Also, don't forget to add c at the end, as a constant.

Also, to the second question, x^4 differentiates to 4x^3

Oh so you mean ∫x^4dx ??