Calculus 1
Question is attached.

- dessyj1

Calculus 1
Question is attached.

- schrodinger

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- dessyj1

Sorry, my internet connection suddenly slows down when i try to upload a picture.

- dessyj1

##### 1 Attachment

- freckles

4d?

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## More answers

- dessyj1

sorry number 5

- freckles

lol oh that is a choice
\[\lim_{h \rightarrow 0}\frac{1}{k}\ln(\frac{2+h}{2})\]

- freckles

I didn't realize it was a multiple choice thingy

- freckles

\[F'(x)=f(x) \\ \int\limits_a^b f(x) dx=?\]
What does F'=f mean?
I will give you a hint that F is the ____-derivative of f.

- freckles

I will also give you another hint:
fundamental theorem of calculus

- freckles

for example:
how do you evaluate this:
\[\int\limits_{1}^{2}x^2 dx\]

- dessyj1

Since i know that the derivative of F(x) is itself i can just switch them around in the integral equation right?

- freckles

recall:
\[\frac{d}{dx}(\frac{x^3}{3})=x^2 \text{ for all } x \\ \ \text{ so } \int\limits_1^2 x^2 dx=[\frac{x^3}{3}]_1^2 =\frac{2^3}{3}-\frac{1^3}{3}\]
you are given
\[\frac{d}{dx}(F)=f \text{ for all } x \\ \int\limits_a^b f dx=[ ? ]_a^b\]
and that f is continuous which is another important thing

- dessyj1

but wouldnt, f(x) in your example be the same function as its derivative?

- freckles

are you saying f=f'?

- freckles

math is case sensitive
so when they say F they don't mean f

- freckles

so no we aren't given f'=f

- freckles

do you know usually to integrate you need to find the antiderivative of the expression that is the integrand ?

- freckles

so if we are given F'=f
that means the antiderivative of f is F
since F'=f

- dessyj1

They do not give is the functions.
but lets assume the function is e^x

- freckles

did you not understand the example I gave above?

- dessyj1

i did not understand it

- freckles

\[\frac{d}{dx}(\frac{x^3}{3})=x^2 \text{ for all } x \\ \ \text{ so } \int\limits_1^2 x^2 dx=[\frac{x^3}{3}]_1^2 =\frac{2^3}{3}-\frac{1^3}{3}\]
I started off exactly as your question did

- freckles

replace the x^2 with f
and replace the x^3/3 with F
you can do this since (x^3/3)'=x^2 and (F)'=f

- dessyj1

the problem with that is the fact that F and f are not the same for all values of x like the question stated.

- freckles

No it is saying F'(x)=f(x) for all x

- freckles

Also why do F and f have to be the same?
You are definitely not given that.

- freckles

F' and f have to be the same for all x

- freckles

which they are because when you differentiate (x^3/3) you do get x^2

- freckles

x^2=x^2 for all x

- freckles

http://mathworld.wolfram.com/FundamentalTheoremsofCalculus.html
This is just the fundamental theorem of calculus

- dessyj1

Okay, I think I have a hard time understanding this because we never learned the fundamental principle of calculus and I am currently studying for the final, so that means my teacher never intended to teach that concept.

- freckles

so you guys never cover definite integrals?

- freckles

covered*

- dessyj1

We did, but we were given the rules.

- freckles

so have you ever done the one or know how to do the one I mentioned before:
\[\int\limits_1^2 x^2 dx\]

- dessyj1

We were not taught how to evaluate an integral using the definition.

- freckles

like how would you tackle that one then?

- dessyj1

I can do definite integrals. Why did you choose x^2? as one of the functions?

- freckles

I can choose 1 or x is doesn't matter
it is just an example

- dessyj1

would e^x work then?

- freckles

\[\int\limits_1^2 1 dx=?\]
sure we can use whatever function that is continuous and has a continuous derivative

- freckles

I just want to see what you do to evaluate something like that if you never been taught the fundamental theorem of calculus

- freckles

Like do you not normally find the antiderivative of the integrand ?

- dessyj1

answer is 1

- freckles

I know but I want to know how you get there

- freckles

like what steps do you take

- dessyj1

ill draw what i did

- dessyj1

|dw:1434604206920:dw|

- freckles

ok good but isn't the derivative of x, 1?

- dessyj1

it is

- freckles

by the way you use the fundamental of theorem of calculus above know it or not

- freckles

(x)'=1 so you are given this

- freckles

and then you said this
\[\int\limits_1^2 1 dx=x|_1^2 \]

- dessyj1

we were that a set of rules to deal with different types of equations such as exponents, and natural logs

- freckles

in our question we are given (F)'=f
so \[\int\limits_1^2 f dx=F|_1^2\]

- dessyj1

Okay

- freckles

I replaced the lower and upper with 1 and 2

- freckles

do you not see this yet \[\int\limits_a^b f(x) dx=F(x)|_a^b=F(b)-F(a) \\ \]?
you know the fundamental theorem of calculus even though you are saying you don't know because you just applied it just a sec ago

- freckles

you know given the other stuff such as f is continuous and F'=f

- dessyj1

this works if you are telling me the integral of f(x) is equal to F(prime)(x)

- freckles

no I'm telling you that at all

- freckles

\[f(x)=1 \\ F(x)=x +C \text{ where } C \text{ is a constant } \\ \text{ do you not agree that } F'(x)=f(x) ?\]

- freckles

Since F'=f
then F is the antiderivative of f

- dessyj1

I do agree, but i feel like you are making the functions up now.

- freckles

\[\int\limits_{a}^{b}f(x) dx \\ \text{ \to integrate this I need the antiderivative of } f \\ \text{ which is } F \\ \text{ since } F'=f \\ \int\limits_a^b f(x) dx=F(x)|_a^b\]
I'm giving you examples

- dessyj1

If were were able to seamlessly communicate this would be easier for me to grasp. But i cannot understand, i will just have to ask my teacher tomorrow. Regardless, thank you for all your help and effort.

- freckles

if f(x)=1
the antiderivative let's call it F is
F(x)=x+c
we will just use F(x)=x since we have a definite integral anyways that we will be working with
\[\int\limits_a^b 1 dx=x|_a^b=(b-a) \\ \int\limits_a^b f(x) dx=F(x)|_a^b=F(b)-F(a)\]

- freckles

like in the example don't you see that (x)'=1
and in the question you are given (F)'=f

- freckles

maybe @ganeshie8 can explain it better
if you think maybe it is just me

- dessyj1

No, i do not think it is just you.

- dessyj1

is the antiderivate of F(prime)(x)= F(x) ?

- freckles

yes!!!

- freckles

\[F'=f \\ \int\limits_a^b f(x) dx=\int\limits_a^b F'(x) dx=F(x)|_a^b\]

- dessyj1

I did not know the mathematical notation for the derivative of a capital function like F(x) was f(x)

- freckles

well
it could have had a different name
like for example maybe they said where g'=f
g'=f still means that g is the antiderivative of f

- freckles

\[\int\limits_a^b f(x) dx=g(x)|_a^b=g(b)-g(a) \text{ since } g'=f \\ \text{ aka since } g \text{ is the antiderivative of } f \]

- freckles

or I think the way you understand it better was like this:
\[\int\limits_a^b f(x)=\int\limits_a^b g'(x) dx \text{ since } f=g' \\ \text{ now } \int\limits_a^b f(x) dx=\int\limits_a^b g'(x)dx=g(x)|_a^b =g(b)-g(a)\]

- freckles

you know assuming g is continuous on [a,b] of course

- freckles

I mean f also

- dessyj1

so the answer to this question is D?

- freckles

yep.

- dessyj1

alright, i get it.

- freckles

don't just say that if you don't believe it
I will not be upset
and @ganeshie8 is here
he is awesome at explaining things if you do not feel I did the trick

- ganeshie8

somehow i feel @dessyj1 you're confusing "anti derivatives" with "definite integrals"

- dessyj1

I do understand the question now. The derivative of F(x) was equal to f(x). That means that the integral(indefinite integral, antiderivative) of f(x) is just like asking for the integral of F(prime)(x) which was F(x)

- dessyj1

i think anti-derivative and indefinite integrals are the same thing.

- ganeshie8

looks perfect!

- dessyj1

Thank you.

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