Question

Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.)
f(x) = x(6 − x)2
for my answer I got X^4/4-4x^3+18x^2+C is this right?

Answer 1

Did you find the derivative to check yourself?

Answer 2

To keep things simple, you may try u-substitution
\(u = 6-x\)

Answer 3

no i didn't find the derivative to check my answer, i would find the derivative of my problem right?

Answer 4

yeah you can differentiate your answer and see if it is x^3-12x^2+36x
which is the form I think you put it in to integrate

(by the way if you have heard of substitutions to make integrals like this nicer looking (or I mean easier) at ganeshie8's note )

Answer 5

nvm if you haven't heard of u-substition before
your work looks good, i presume you have used the formula for antiderivative of \(x^n\)


to double check, differentiate the answer as freckles said

Answer 6

he might heard of it
I don't know

Answer 7

Im not sure i have, but i took the derivative and got x^2-12x+36x :)

Answer 8

x^3-12x^2+36x?

Answer 9

yeah its just that i had x(x^2-12x+36)

Answer 10

and yes I already checked your answer integrating thingy
but your problem said to also check your solution by differentiating
so now you done both things your problem asked for

Answer 11

got it right! thanks!!

Answer 12

how do you know ` x(x^2-12x+36) ` is same as the original function ?

Answer 13

that might be a dumb q hmm

Answer 14

x(6-x)^2
think he expanded the square thingy first

Answer 15

she* and yeah i did

Answer 16

Ahh looks great she :D

Answer 17

oops sorry

Answer 18

do you want to see the sub thingy mentioned by ganeshie8 for fun?

Answer 19

Its okay! sure is it easier than finding the derivative?

Answer 20

\[\int\limits x (6-x)^2 dx \\ \text{ Let } u=6-x \\ \frac{du}{dx}=0-1 \\ \frac{du}{dx}=-1 \\ du=-1 dx \\ \text{ multiply both sides by -1 } \\ -du=dx \\ \text{ now recall if } u=6-x \text{ then } x=6-u \\ \text{ so we have } \\ \int\limits x (6-x)^2 dx=\int\limits (6-u)u^2 (-du) \\ =\int\limits (u-6)u^2 du\]
so now you have less multiplication to do

Answer 21

\[=\int\limits (u-6)u^2 du =\int\limits (u^3-6u) du=\frac{u^4}{4}-3u^2+C \\ \text{ now remember } u=6-x \\ \text{ so you have } \\ =\frac{(6-x)^4}{4}-3(6-x)^2+C\]
and it is less multiplication if you get to leave your answer like this :)

Answer 22

\[
x(6 − x)^2 = x(x^2-12x+36) = x^3-12x^2+36x
\]Then use power rule. No tricks needed. Your anti-derivative looks correct.

Answer 23

Oh okay, I didnt understand the substitution part at first, but i get it now! thanks for explaining!

Answer 24

oops I didn't multiply correclty

Answer 25

my answer is off
because when I did -6(u^2) I put -6u
and so I integrated the wrong thingy

Answer 26

but that is sorta how works above if you can multiply correctly :p

Answer 27

you should ask why on earth substitution is any better than your original method @gaba

Answer 28

try this if you're loving antiderivatives :
\[\int x(6-x)^{9999999}\,dx = ?\]

Answer 29

I'm deff not loving them lol
so Im working on this right now, Find the most general antiderivative of the function 2rootx+6cosx
I got 4x^1/2+6-sinx

Answer 30

im not sure i did it right though

Answer 31

differentiate your answer and see if you get back the original function

Answer 32

also what happened to the constant, C

Answer 33

nvm I just did it again and got 4x^3/2/3 +6 -sinx +C

Answer 34

you're doing calculus, that means you're not in highschool anymore
time to use proper notation

Answer 35

4x^ `3/2/3` +6 -sinx +C

what does that even mean

Answer 36

[(4x^3/2)/3] +6-sinx+C
Btw you can take calculus in high school, just saying

Answer 37

Much better, but still it is not mathematically correct

Answer 38

pretty sure you meant

[(4x^3/2)/3] +6(-sinx)+C

Answer 39

differentiate that and see if you get back the original function

(there is a mistake, so you wont get back the original function)

Answer 40

why is it +6(sin(x)) and not -sin(x)?

Answer 41

\[f(x)=2 \sqrt{x}+6 \cos(x) \\ \text{ or we can write } \\ f(x)=2 x^\frac{1}{2}+6 \cos(x)\]
is this what you are playing with?

Answer 42

yes!

Answer 43

anyways
( ? )'=cos(x)
the ?=sin(x) right?

(sin(x))'=cos(x)
so the antiderivative of cos(x) is sin(x)+C


if I had
( ? )'=sin(x)
the ?=-cos(x)
(-cos(x))'=sin(x)
so the antiderivative of sin(x) is -cos(x)+C



now if you had
( ? )'=-cos(x)
then ?=-sin(x)
(-sin(x))'=-cos(x)
so the antiderivative of -cos(x) is -sin(x)+C




anyways you had
find the antiderivative of 2x^(1/2)
which looks like you were going for: 4x^(3/2)/3 +k

but to find the antiderivative of 6cos(x)
you need to recall what can you take derivative of that will give you 6cos(x)
you can bring down the 6( and put the antiderivative of cos(x)) here )

Answer 44

forgot to put + some constant at the end there

Answer 45

oh okay, is that to make sure I had it right I checked on mathway for the derivative of cos(x) and it said it was -sin(x) (which is what I thought it was before checking so it made sense) but i typed in +sin(x) and got it right, so thanks for explaining everything!

Answer 46

oh so you got derivative and antiderivative mixed up

Answer 47

but just so you know if it was taking derivative of 6cos(x)
you would put -6sin(x) and not 6-sin(x)

Answer 48

but yeah antiderivative of 6cos(x) would be 6sin(x)

since the derivative of 6sin(x) is 6cos(x)

Answer 49

the antiderivative of 6cos(x) would be 6sin(x)+C
since the derivative of 6sin(x)+C is 6cos(x)
*

Answer 50

the most general antiderivative *

Answer 51

Okay I understand now! thanks a lot!!

Answer 52

np