Question

Integral? (Antiderivative) of x^(2/5)

Answer 1

Not sure on the terminology :-)

Answer 2

you function to be intrgrated, is of the form x^n , where n = 2/5
so you will use
\(\large \int x^ndx = \dfrac{x^{n+1}}{n+1}+c\)

Answer 3

Sooo x^(2/5)+1=x^7/5/(7/5)+c?

Answer 4

antiderivative 12x^2-4x+3

Answer 5

that 5 x^(7/5)/7 +c was correct :)
now we use the fact that integral of sum is sum of integrals.
so,
\(\int12x^2dx-\int 4xdx +\int3dx \)
and constants can be pulled out of the intrgral

Answer 6

=\(12\int x^2dx-........\)

Answer 7

use same formula of x^n
treat 3 as 3x^0 (that is n=0)

Answer 8

Exactly the same? so just for 12x^2 I'm going to do (12x^3)/3+C and move on?

Answer 9

absolutely! :)

Answer 10

That's pretty easy..

Answer 11

Okay wait now I have a question. What is the full answer? I just want to see the form you're writing it in

Answer 12

because 12x^3/3 is 4x^3 and do I use three +C's? so that my final answer would be +3C?

Answer 13

yes, most of the integration problems are easy, with few exceptions ofcourse, especially when you have just started it :) with enough practice, you can master the integration thing :)

\(\large \dfrac{12x^3}{3}-\dfrac{4x^2}{2}+\dfrac{3x}{1}+c \\ \large = 4x^3-2x^2+3x+c\)

Answer 14

So only use one +c, so exactly like the other formula just no C until the end

Answer 15

\[\int\limits_{}^{}\sqrt{x^{11}dx}\]

Answer 16

that + c is just there to denote, that there can be a constant after the integration process. you can get the exact value only if you have some initial conditions.
so whether you write c or 3c or c1 ....it doesn't matter :)

Answer 17

can you write that in the form of x^n ?
n=..?

Answer 18

This one must be different because technically is x^(11)^1/2

Answer 19

why would it be different ? isn't it in the same form of x^n ?

Answer 20

x^11/2?

Answer 21

x^11^1/2 can be written as ...?
yes, its x^11/2 :)
so just put n= 11/2 !

Answer 22

x^11/2+1=x^13/2/(13/2)+C

Answer 23

don't write +1 on left side
\(\int x^{11/2}dx = \dfrac{2x^{13/2}}{13}+c\)

Answer 24

HMM then what about for 7/x-4/x^3?

Answer 25

If this one isn't different.. lol :(

Answer 26

1/x can be written in the form of x^n ?

Answer 27

x^-1

Answer 28

and
\(\dfrac{1}{x^3} = x^{???}\)


oh and the first part is indeed different

Answer 29

\(\huge \int \dfrac{1}{x}dx = \ln |x|+c\)
because differentiation of ln x is 1/x

Answer 30

x^-3

Answer 31

correct
so for the 2nd part you use x^n formula with n = -3

Answer 32

7x^-1-4x^-3? Lol :(

Answer 33

yes, thats your question
you still need to integrate that, right ?

Answer 34

for the first part use the new formula i gave for 1/x
for the 2nd part , use x^n formula with n=-3

Answer 35

OOO okay I thought that was the final answer I was like that doesn't seem like much of a change, but okay I get it.

Answer 36

yeah so integrate it and tell me what u get ?

Answer 37

7x^0/0-4x^-2/-2+c

Answer 38

why did you use x^n formula for 1/x when i gave you a new formula for it ?
the 2nd part "-4x^-2/-2" is correct! just simplify it

Answer 39

the reason you cannot use the x^n formula for 1/x is that you get a 0 in the denominator, which is not allowed.

Answer 40

I don't see why you can use that, you said 1/x so I figured it had to be 1/x to use it?

Answer 41

\[\frac{ 7 }{ x }-\frac{ 4 }{ x^3 }\]

Answer 42

thats the question

Answer 43

I GOT IT i see

Answer 44

so whats your final answer now ?

Answer 45

Your formula kind of confused me because you put 1/x and I was like, how exactly do you get 1/x? But I see you take the top term and put it on the outside of the antiderivative so it's \[7\int\limits_{}^{}\frac{ 1 }{ x }\]

Answer 46

7ln(x)+2/x^2+C

Answer 47

the top term(7) was a CONSTANT.
that is why we can bring it out of the integral
if it were variable, then we could have not brought it out of integral

Answer 48

and your final answer is correct :)

Answer 49

Sorry I barely know my terms lol, just know how to do the work :(

Answer 50

\[4e^{-7x}dx\]

Answer 51

ok, since derivative of e^x is just e^x,
\(\int e^x dx = e^x+c\)

Answer 52

but we have the exponent as -7x here
so, we make a substitution

Answer 53

let -7x = u
we need to find dx now in terms of 'u' and 'du'
can you differentiate this ?
-7x = u
what do u get ?

Answer 54

Sorry I solved through, -4e^-7x/7+C

Answer 55

thats correct, did you understand how to get that ?

Answer 56

Yes :-)
Find f such that f'(x)=5x^2+3x-5 and f(0)=8

Answer 57

since we know that integral is anti-derivative
the question is just
\(\int 5x^2+3x-5 \:\: dx \)
right ?
and i think you can solve this integral

Answer 58

tell me what u get after integrating, then i'll tell you how to get the value of c

Answer 59

integral:
5x^3/3+3x^2/2-5x+C

Answer 60

yes, so your f(x) = 5x^3/3+3x^2/2-5x+c
is correct :)
now we are given f(0) = 8
so to get f(0), just put x = 0 in f(x)
what do u get ?

Answer 61

8..? lol

Answer 62

what u get if you put x= 0 in
f(x) = 5x^3/3+3x^2/2-5x+c
?

Answer 63

did you mean c= 8 ?

Answer 64

Yes C=8

Answer 65

then you are correct

Answer 66

f(x) = 5x^3/3+3x^2/2-5x+8
is the final answer.

Answer 67

Last question :-)

Answer 68

\[\int\limits_{}^{}\frac{ u^{5}-2u^{3}+8 }{ 3u }du\]

Answer 69

separate out the denominator

Answer 70

like (a+b+c)/d = a/d+b/d+c/d

Answer 71

u^5/3u-2u^3/3u+8/3u

Answer 72

yes, so whats
u^5/3u = ... ?

Answer 73

Now it's just applying everything I did before lol

Answer 74

yes

Answer 75

u^3/3-2u-4/u^2+C

Answer 76

what have you done :O

Answer 77

All of it!! :D

Answer 78

go term by term :P then you won't do any error
u^5/3u = ... ?

Answer 79

Well it marked right :P

Answer 80

LOL!
wheen it clearly was incorrect :P

Answer 81

What?!:( What was wrong? haha

Answer 82

i thin i am asking you this for 3rd time
u^5/3u =... ?

Answer 83

u^4/3

Answer 84

WAIT SO HOW DID I GET IT RIGHT XD

Answer 85

thats your 1st term to integrate
so, you get
(1/3) u^5/5
after integration, right ?

i have no idea how you got that right :P

Answer 86

Let me check my work idk

Answer 87

Honestly I did an error but MathWay is showing u^3/3 and wolfram is showing something way different.. hrmm lol

Answer 88

it should be 4.. wth LOL

Answer 89

when you can do it on your own, why go to virtual answering sites ? :P

Answer 90

To check my answer for questions like why this is marked right when its wrong. :O

Answer 91

what should be 4 ?

Answer 92

= u^5/3u-2u^3/3u+8/3u
= u^4/3 -2/3 u^2 +8/3 (1/u)

taking the integration
= 1/3 (u^5/5) -2/3 (u^3/3) + 8/3 ln |u| +c
see if u get this