Question

\[If~~ f(x) =\int\limits_{0}^{x}(t ^{2}-3t+2) dt\]find f'(x)
I got
\[f'(x) = \frac{ 1 }{ 3 }x ^{3} - \frac{ 3 }{ 2 }x ^{2}+2x\]would that be correct? or am I totally wrong? XD

Answer 1

No silly billy that's not right! D:

You correctly calculated \(\Large\bf\sf f(x)\)
Now take its derivative.

Answer 2

so x^2 - 3x + 2 ?

Answer 3

Yay good job \c:/

Answer 4

Awesome :D
Thanks!

Answer 5

You could have also applied the `Fundamental Theorem of Calculus: Part 1` to this problem:\[\Large\bf\sf \frac{d}{dx}\int\limits_0^x f(t)\;dt \quad=\quad f(x)\]

Answer 6

But going through it the long way is good, you get familiar with the process c:

Answer 7

coolios :)
I have no idea what I'm doing XD my teacher says there's a short cut? :3

Answer 8

Yah that's the shortcut :U

Answer 9

Example Problem:

Answer 10

\[\Large\bf\sf f(x)\quad=\quad \int\limits_0^x \frac{e^{\arctan(\sqrt[3]{t^2-1})}}{3t^2+7t-1}dt\]
Then the derivative is simply,\[\Large\bf\sf f'(x)\quad=\quad \frac{e^{\arctan(\sqrt[3]{x^2-1})}}{3x^2+7x-1}\]No need to do all the fancy work of trying to find an anti-derivative and then differentiate it again.

Answer 11

ohhh so you just replace t with x ?

what if it was 0 to x^2 ?

Answer 12

Ohh good question.
Then when you go to differentiate, you have to apply the chain rule.\[\Large\bf\sf \frac{d}{dx}\int\limits\limits_0^{x^2}f(t)\;dt\quad=\quad \frac{d}{dx}F(t)|_0^{x^2}\]
\[\Large\bf\sf =\frac{d}{dx}\left[F(x^2)-F(0)\right]\]Differentiationg finally,\[\Large\bf\sf =f(x^2)\frac{d}{dx}x^2-0\quad=\quad f(x^2)\cdot 2x\]

Answer 13

So as before, we simply stick x^2 in for all of the t's.
But now we also get an extra factor of 2x because of the chain rule.

Answer 14

mmm then how would trig work?

cuz I was given
\[f(x) = \int\limits_{0}^{x ^{2}}(t + \cos 3t)dt\]

Answer 15

Our integrand is a function,
\[\Large\bf\sf f(\color{royalblue}{t})\quad=\quad \color{royalblue}{t}+\cos3\color{royalblue}{t}\]We take it's anti-derivative, giving us some function we'll call,\[\Large\bf\sf F(\color{royalblue}{t})\]Plug in the boundaries, and take our derivative,\[\Large\bf\sf =\quad f(\color{royalblue}{x^2})\cdot 2x\quad=\quad (\color{royalblue}{x^2}+\cos3\color{royalblue}{x^2})\cdot 2x\]

Answer 16

It's important to understand what is going on.
But for the sake of simplification you can think of it as,
~I get the same thing back, but with `the upper bound` inside the function instead of t,
~and i have to multiply by the derivative of the upper bound.

Answer 17

omg that short cut is so cool :D

if you don't mind helping me with one more? :3

\[\huge f(x) = \int\limits_{-1}^{xsinx}\sqrt{t ^{2}+1}dt\]

Answer 18

Do you understand why the lower boundary is giving us zero?
(It will happen in this problem as well)

Answer 19

mmm before when lower bound was 0
it wasn't involved cuz it was stuff times 0 which equals 0 right?
but now we has a -1 as lower bound
so would it be like F(xsinx) - F(-1) ? (has no idea what that notation even means XD)

Answer 20

So like.. think of a simple example really quick,
Because the notation is important. :)

\[\Large\bf\sf \int\limits_a^x f(t)\;dt\]We have some stuff inside that we're integrating.
We're generalizing by calling it a function.
We use big f to denote the anti-derivative.

We might not actually be able to find the anti-derivative but that doesn't matter.
We just call it F(t).


The process is,
~Anti-derivative
~plug bounds in
~take derivative

If you integrate something, then differentiate it, you should expect to get back what you started with, yes?

It's like you're doing something, then you're undoing it.
But with this process we do something between the doing and undoing.
We change the argument of our function.

Answer 21

So when you integrate you get some anti-derivative function,\[\Large\bf\sf F(t)\]Then evaluating it at the upper and lower bounds,\[\Large\bf\sf F(x)-F(a)\]Now when we take a derivative (with respect to x),
the second piece that we have there....... is just a constant.

Answer 22

It's some function (that we're pretending we found),
and we're evaluating it at a constant.

Answer 23

The derivative of a constant is zero, right?

Answer 24

oh right yea :)

so it would still be the same idea? what would happen to the -1?

Answer 25

We'll call our integrand \(\Large\bf\sf f(t)\).
So let's assume that some anti-derivative exists.
Integrating gives us,\[\Large\bf\sf F(t)\]Evaluating it at the upper and lower bounds gives us,\[\Large\bf\sf F(x \sin x)-F(-1)\]as you said.

Now we need to take a derivative.

Answer 26

If I have the function,\[\Large\bf\sf G(x)=x^2\]And evaluate it at -1,\[\Large\bf\sf G(-1)=1\]What is the derivative of G(-1)?

Answer 27

zero?

Answer 28

Yes, you simply took the derivative of 1, correct?

Answer 29

yeah :D

Answer 30

Whenever we take the derivative of a function which has a `constant` plugged in, then it just gives us zero.

Answer 31

\[\Large\bf\sf \frac{d}{dx}\left[F(x \sin x)-\color{orangered}{F(-1)}\right]\quad=\quad f(x \sin x)\frac{d}{dx}(x \sin x)-\color{orangered}{0}\]

Answer 32

We get back the same thing we started with, but with (x sin x) plugged in for all the t's.
And we also have to multiply by the derivative of (x sin x).

Answer 33

which is xcosx + sinx ?

Answer 34

Mmmm yah looks good.

Answer 35

\[\Large\bf\sf\quad=\quad \sqrt{(x \sin x)^{2}+1}\left[x \cos x + \sin x\right]\]

Answer 36

should I simplify or is than an acceptable answer? :3

Answer 37

That's a good answer. Don't mess with it :)

Answer 38

okay thank you! :D

I'll try to do the rest myself, and if I still don't understand I'll ask :)
Thank you sooo much! <3

Answer 39

np ^^

You might run into another form,\[\Large\bf\sf \frac{d}{dx}\int\limits_x^0f(t)\;dt\]Something like that perhaps, where the variable is in the lower limit.
You'll want to get it into the upper limit before you can apply this shortcut.

Answer 40

right XD (looking at one now) it'll just be a negative in front right? :)

Answer 41

Yes good!
That's how you switch the bounds :)

Answer 42

so \[f(x) = \int\limits_{x^2}^{4} sint ~dt\]becomes
\[f'(x) = -2xsin(x^2 +1)\]

Answer 43

whoops lower bound x^2 + 1

Answer 44

Oh ok, ya looks correct c:

Answer 45

awesome :)

and just to check if I got the concept down...
\[\large f(x) = \int\limits_{3-x}^{2x}\sec^5(t) dt\]would be \[\large f'(x) = 2\sec^52x + \sec^5(3-x)\]

Answer 46

So normally we would `subtract` the lower bound.
But it looks like you got an extra negative from the chain rule, so you just combined them.
Yessss good job!

Answer 47

YAYYYY!!!!
:D
*happy dance*

Answer 48

hah XD