Question

let \[f(x) = \sqrt[3]{x^2+4x}\]
and let g(x) be antiderivative of f(x). then if g(5) = 7 then find g(1)

Answer 1

let \[f(x) = \sqrt[3]{x^2+4x}\]
and let g(x) be antiderivative of f(x). then if g(5) = 7
then find g(1)

Answer 2

rutn it into an exponenetial form

Answer 3

hmm

Answer 4

\[f(x) = ({x^2+4x})^{1/3}\]

Answer 5

turn it into

Answer 6

then power rule it, but always include the chain rule that pops out with it

Answer 7

hmm k give me a sec

Answer 8

\[f(x) = ({x^2+4x})^{1/3}\]
\[g(x) = \frac{1}{3}({x^2+4x})^{-2/3}\ *[x^2+4x]'\]
\[ g(x)= \frac{1}{3}({x^2+4x})^{-2/3}\ (2x+4)\]
\[ g(x)= \frac{2x+4}{3(x^2+4x)^{2/3}}\]

but thats just me deriving it .... you want the integration ...

Answer 9

what about it : g '(x) = f(x) and then \[g(x) = \int\limits_{a}^{x}\sqrt[3]{t^2 + 4t}dt\] then we let h(x) = g(x) - 7 and \[h(x) = \int\limits_{5}^{x}\sqrt[3]{t^2 + 4t}dt\]

Answer 10

yeap i want the integration :D

Answer 11

maybe integration by parts?

Answer 12

mybe :) so you gonna do that? \[h(r) = \int\limits_{5}^{x}\sqrt[3]{t^2}dt + \int\limits_{5}^{x}\sqrt[3]{4t}dt\]

Answer 13

lol .... maybe not

http://www.wolframalpha.com/input/?i=int+%28x^2%2B4x%29^%281%2F3%29+dx

Answer 14

turn it into a power series if need be .... and play with the poly that you get

Answer 15

if you think its gonna be an esay split; it aint, not if the wolf spits back that thing

Answer 16

hmm

Answer 17

i think i just did a simple mistake on using integration by parts o.O , i forgot a rule i guess you can't use it in roots! ??

Answer 18

this integral sucks. there must be some other gimmick at work here to solve the problem without using integration

Answer 19

o.O hmm

Answer 20

i see no snap way to answer this question. maybe i am missing something

Answer 21

well i think my way solves the question mybe? last i said that \[h(x) = \int\limits_{5}^{x}\sqrt[3]{t^2 + 4t}\] and if we will solve it for h(1) (h(x) is an antiderivative of f(x) .... we solve that and we got (calculator <3 <3 ) -10.88222 since h(x) = g(x) - 7 , g(x) = h(x) + 7 and i have to go to dinner brb

Answer 22

at the moment( h(x) = g(x) - 7 )

Answer 23

oh ok now back
give me a minute

Answer 24

and it fallows that : g(x) = h(x) + 7

then we can do that g(1) = h(1) + 7 = -10.88222 + 7 = -3.88222 !!!!!!!!!!

Answer 25

err is this true what i did?

Answer 26

so if we look what we did :
take g(x) as antiderivative of f(x)
than we will get g'(x) = f(x)
or \[g(x) = \int\limits_{a}^{x}\sqrt[3]{t^2 + 4t}dt\]
but its unsolveable so instead of integrating we let
p(x) = g(x) - 7
(p(x) is also an antiderivative of f(x) and thann we can write
\[p(x) = \int\limits_{5}^{x} \sqrt[3]{t^2 + 4t}\]

!!! now by using calculator we easly got
\[p(1) = \int\limits\limits_{5}^{1} \sqrt[3]{t^2 + 4t}\] = 10.88222 (CALCULATOOOR <3!!)
we said p(x) = g(x) - 7
g(x) = p(x) + 7
so that g(1) = p(1) + 7 = -10.88222 + 7 = -3.88222

Answer 27

i guess its right :D

Answer 28

\[\int\limits_{5}^{x}f(t)dt=g(x)-g(5)\]

\[g(x)=\int\limits_{5}^{x}f(t)dt+7\]


so \[g(1)=\int\limits_{5}^{1}f(t)dt+7=-\int\limits_{1}^{5}f(t)dt+7\approx -3.88222\]

Answer 29

hold the phone
where did the number come from???

Answer 30

which number? you mean limits?

Answer 31

no the 10.333 whatever?

Answer 32

-10.88 etc

Answer 33

use calculation to solve the integral and you get that result :D

Answer 34

ooooooooooooooh. i thought it was some theory you were supposed to use! numerical integration. ok

Answer 35

and i used second fundemental theorem of calculus here :D

Answer 36

lol

Answer 37

lol :)

Answer 38

you used ti 89, wolfram, whatever

Answer 39

hello

Answer 40

yeap i just made the integral as solvable, found limits :D used second fundemental theorem of calculus than used a calculator to solve the integral :D :D

Answer 41

hello!
integrate from 1 to 5 and adjust. cheating if you ask me...

Answer 42

Ahaha man com'on :D its the solution anyway

Answer 43

hi imranmeah!

Answer 44

no i like it it is good for sure

Answer 45

3 mounths ago, i did not even know what limits are :P

Answer 46

(i mean i did not kno derivatives, lineer algebra, 0 knowledge of algebra and calculus) :D

Answer 47

you can get bounds for the solution by using the mean value theorem...But I see no nice way of getting the exact answer.

Answer 48

u right, but i'm not sure of it

Answer 49

that is for sure!