Question

what is this??
I am very confused

Answer 1

you have to use u-substitution. So u could be 1+3x^4

Answer 2

ok,
so
u=1+3x^4?
du=12^3

whats a nd b?

Answer 3

a and b are modified limits. To evaluate them you substitute 'em for x into whatever u was.

Answer 4

So for a=2 --> 1=3(2)^4=49

Answer 5

so a isnt just 2
nd b inst 4?

Answer 6

no it's 49 and something else

Answer 7

how did you get 49?

Answer 8

u=1+3x^4
du=12x^3dx
a=769
b=49
f(u)=u^2
Value of the integral=1268860

Answer 9

if u = 1+3x^4 and the bottom limit is 2, then a = 1+3(2)^4 = 49

Answer 10

Yes...you are right...I wrote them the other way around....sorry about that...

Answer 11

Did it make sense?

Answer 12

Oh I see i got it.
thanks
but f(u) and intergals were wrong

Answer 13

The computing the limits was always (is) confusing for me...just remember that if you use u substitution then you can either re-calculate the limits in terms of u or you can put the whole integral back in therms of x after you are done with substitution. If you decide to put the integral back in terms of x then you can get away with re-calculating the limits....it is your choice really...

Answer 14

What was the answer for f(u)? If you don't mind me asking?

Answer 15

u^2???

Answer 16

(1+3x^4)^2=(1+3x^4)(1+3x^4)=1+6x^4+9x^8

Answer 17

1+6x^4+9x^8 is the answer?

Answer 18

@cblrtopas

Answer 19

\[\int _2^4(1+3x^4)^2x^3dx\]
\[u=1+3x^4, du =12x^3dx\]

Answer 20

\(u(2)=1+3\times 2^4=49, u(4)=1+3\times 4^4=769\)

Answer 21

that means \(a=49,b=769\)

Answer 22

yes i got these above.. but not f(u) and the rest of them are not solved

Answer 23

then since \(du=12x^3dx\) and since you only have \(x^3dx\) you need
\[\frac{du}{12}=x^3dx\]

Answer 24

that makes your integral
\[\frac{1}{12}\int_{49}^{769}u^2du\]

Answer 25

I see! how do i solve next? finding replace u right?

Answer 26

anti derivative of \(u^2\) is \(\frac{u^3}{3}\) plug in your numbers etc

Answer 27

ok so once i set the equation,
i need to find u which needs to be anti derivatived.
then plug number into u?

Answer 28

i am not sure exactly what you mean in that question
your job is now to compute \[\frac{1}{12}\int_{49}^{769}u^2du\]which you do by taking the anti derivative of \(u^2\) which is \(\frac{u^3}{3}\)
evaluate at the upper limit, evaluate at the lower limit, and subtract
also divide by that other \(12\) outsides

Answer 29

literally your answer is
\[\frac{1}{36}\left(729^3-49^3\right)\]

Answer 30

I think i got it . i will try other questions to see if i understand fully
thank you very much

Answer 31

yw