Question

Int x^4(x5 - 8)^4 dx [0,5]
lost on this one, could use some pointers

Answer 1

use (x5 - 8)=z and then u will get the result

Answer 2

then find dz which is the dirivitive right?

Answer 3

dz=5x^4?

Answer 4

yup dz=5x^4dx

Answer 5

even change the limit if u want otherwise u can simplify the integration then bring it back in terms ox and then substitute

Answer 6

ok I think im missing something...\[\int\limits_{?}^{?}x ^{4}(u ^{4})dx\]
then
\[1/5 x ^{5}(u ^{4})\]

Answer 7

do I u sub again for the u^4?

Answer 8

it is 1/5integral of (u^4) du =(u^5)/25
now bring the answer iback in x and the nsubstitute for x

Answer 9

\[1/5 x^51/5u^5=1/25 x^5u^5?\]

Answer 10

sorry, I'm having a hard time with this one

Answer 11

why u r writing x^5 always it seems u hv not cancelled it with x^4

Answer 12

so it is just 1/25 (u^5)?

Answer 13

Sorry, lost again

Answer 14

can you write it down properly with the integrals?

Answer 15

using formula?.. : /

Answer 16

\[\int\limits_{?}^{?}x^4(x^5-8)^4\]

Answer 17

Evaluate the integral

Answer 18

Tried to u sub for (), but keep getting lost

Answer 19

I think I got it...

Answer 20

\[(x^5-8)^4/25\]

Answer 21

\[(x^5−8)^5/25\]

Answer 22

i havent seen those for a few years but you can always find the (a-b)^4 and just simplify it until the end...

Answer 23

just a lot of work... is it supposed to be easy exercise ?

Answer 24

not really. Usually are a bit challenging

Answer 25

I have another that I just need a sanity check on...

Answer 26

then just solve ιt with the time consuming way

Answer 27

you will need this:
(a-b)^4 = a^4 - 4a^3b+6a^2b^2-4ab^3+b^4

Answer 28

..and just do the calculations

Answer 29

\[\int\limits_{?}^{?}1/(16-x^2)^2 dx =(16-x^2)^{-1}+c\]

Answer 30

u=16-x^2
du=-xdx

Answer 31

\[-\int\limits_{}^{}u^{-2}du\]

Answer 32

about the second i think you re right

Answer 33

Hope so. Thanks