Question

antiderivative of \[\int\limits_{0}^{1} (x^4 + 5)^3 dx\]

Answer 1

please help me

Answer 2

expand the integrand.

Answer 3

what do you mean? substitutions?

Answer 4

Unfortunately with this one, you don't have a nice clean substitution available. You'll need to expand out the binomial. (Multiply out the cube thingy) :D

Answer 5

one sec thats not the full problem

Answer 6

\[\int\limits_{0}^{1}x^3(x^4 + 5)\]

Answer 7

Oh lol

Answer 8

lol sorry

Answer 9

Soooo do you see something + it's derivative? :O

I see x^4 and an x^3 term.. hmm looks like we can do a nice U sub again.

Answer 10

u= x^4 + 5
u= dx^3
1/4 u= x^3 dx

Answer 11

??

Answer 12

\[u=x^4+5\]\[\frac{ 1 }{ 4 }du=x^3 dx\]

Answer 13

yea i forgot the du

Answer 14

next, is it 1/4 antiderivative U^3 dx = 1/4

Answer 15

U^4/4 | 6/5 = 6^4 - 5^4/ 16

Answer 16

is it confusing to you?

Answer 17

no you need to set up the integral correctly :D slow down a sec

replace all of the X's and DX with U's and DUUUUUU
you still have a dx for some reason in your integral.

Answer 18

1/4 antiderivative u^3 du <- this shouldn't be a dx anymore.
Otherwise yes it looks ok.

Answer 19

so apart from that everything else is right?

Answer 20

yes, it looks like you found your new limits for U, so thats good.
and plugged them in correctly. yay.

Answer 21

thanks i have got two questions left

Answer 22

\[\int\limits_{0}^{1}e^(4T + 3) dT\]

Answer 23

the (4T + 3) are exponents

it is supposed to be e^(4T + 3) dT

Answer 24

we have to do some subs again right

Answer 25

We don't have to, but yes we probably should. :) So let's do that just so you can get the rhythm of what is going on with exponentials.

Answer 26

u=4T + 3
du= 4dT
1/4du= dT

Answer 27

yah looks good.

Answer 28

1/4\[\int\limits_{5}^{7}e^4du=1/4 \]

Answer 29

e^4| 7/3= e^7- e^3/4

Answer 30

why do your u's look like 4's? i don't understand :o

Answer 31

huh

Answer 32

e^4...?

Answer 33

and also, i think your limits should be from 3 to 7, if im calculating that correctly :o

Answer 34

when t=0, u=5
when t= 1, u=7

so its supposed to be e^7- e^5/4 right?

Answer 35

when t=0
u=4*0+3
u=3

right?

Answer 36

\[\huge \int\limits_0^1 e^{(4T+3)}dT \rightarrow \qquad \frac{1}{4} \int\limits_3^7 e^u du\]

Answer 37

your right

Answer 38

next step would be
1/4 e^4| 7/3

yea?

Answer 39

what is e^4? why isn't it e^u?

Answer 40

right, i keep making that mistake

Answer 41

apart from this e^4 mistake, everything else is fine?

Answer 42

yah looks good.

Answer 43

thanks last but not least

Answer 44

the last response you gave me in this thread, was that suppose to be the final answer?

http://openstudy.com/users/lukasviktor#/updates/50a83c05e4b0129a3c9022f6

Answer 45

no we didn't apply the upper and lower limits on it yet

Answer 46

can we finish it please