Question

Integration Help!

Please guide through the steps
\(\huge \sf \int\limits_{0}^{4} (6t-8)(3-8t+3t^2).dt\)

Answer 1

ok 1st we need to open the brackets :)

Answer 2

wait....the method with which i did it was really long but i got the answer 176 wait lemme think of a shorter one :)

Answer 3

hmmm. I can do this...
but as @imqwerty is doing.
I don't need to interfere.
If he has trouble then I will jump in @Abhisar

Answer 4

u substitution

Answer 5

You mean like this

\(\sf ∫u v dx = u∫v dx −∫u' (∫v dx) dx\)

Answer 6

Well, we know that if you have something like an abstract case below:


\(\large\color{black}{ \displaystyle \int_{a}^{b} \color{red}{f'(x)}~{\rm G}\left(\color{red}{f(x)}\right)dx }\)

f'(x) is just some other function that is derivative in relation to the f(x) part,

then you set \(u=f(x)\)
\(du=f'(x)~du\)
\(x=a~~~\rightarrow~~u=f(a)\)
\(x=ba~~~\rightarrow~~u=f(b)\)

and you would then have:
\(\large\color{black}{ \displaystyle \int_{f(a)}^{f(b)} {\rm G}\left(\color{red}{u}\right)du }\)

Answer 7

I meant \(du=f'(x)~dx\)

Answer 8

should i give ordinary, reg. example?

Answer 9

Yes, please c:

Answer 10

\(\large\color{black}{ \displaystyle \int_{2}^{4} 2x\cdot (x^2-3)^{100}dx }\)

you don't want to epand, do you? O~O

you set \(u=x^2-3\)
and then
\(du=2x~dx\) (see how nice it is that you have that "2x" part in the integral)

and when x=2, u=\((\color{red}{2}^2-3)=1\) (because u in relation to x is x²-3)
and when x=4, u=\((\color{red}{4}^2-3)=13\)


So you get:

\(\large\color{black}{ \displaystyle \int_{1}^{13} (u)^{100}du }\)

Answer 11

and then the rest in that example is obvious (note, that I change the limits of integration equivalently, so you don't need to sub back the x)

Answer 12

I did it in a different way..... (I dont understand why it should be wrong)


I multiplied the brackets

Answer 13

Please, wait....Let me digest it c:

Answer 14

take your time