Question

x^2(5x^3-2)^3 dx on [2,0]
find the exact value of the definite intergral

Answer 1

hmm this is pretty straightforward \[\huge \int_0^2 x^2 (5x^3 - 2)^3dx\]
let u = \((5x^3 - 2)\)
du = \(15x^2dx\)
\[\huge \implies \frac{1}{15} \int _0^2 u^3 du\]
does that help?

Answer 2

yes a little.

Answer 3

well now use power rule

Answer 4

then plug in the limits and you can be on your merry way

Answer 5

@hooverst lgbasallote did good work ..just forgot to change limits as well
so
when
u=5x^3-2

when x=0
then
u=5(0)^3-2=-2
and
when
x=2
then
u=5(2)^3-2=38
so the integral becomes
\[\Large \frac{1}{15}\int\limits_{-2}^{38}{u}^3du\]

Answer 6

hmm...i really dont care about the limits =_= i tend to solve for u..change back to x then just plug in limits..it's less confusing lol

Answer 7

gud .. but it gets complex little bit the n!! \[\Large \frac{(5x^3-2)^4}{4}\] then limits omg ! :P
what about just
\[\Large {u^4}\] and then limits

Answer 8

you have your preferences i have mine :p

Answer 9

ok :P