Question

How do I evaluate this integral?

Answer 1

Okay, it says to Find \[\prime(x)=\int\limits \left(\begin{matrix}2013 \\ 5-x^2\end{matrix}\right)\sqrt{10-t} dt\]

Answer 2

any help would be really appreciated! :(

Answer 3

trying to find

Answer 4

\[f \prime(x)=\]

Answer 5

\[\Large f'(x)\quad=\quad \int\limits_{5-x^2}^{2013}\sqrt{10-t}\;dt\]Find \(\Large f(x)\).

Is that the problem?
The way you wrote it is a little strange, so just checking :o

Answer 6

yes, but switch the f(x) with f '(x). sorry I didn't make it clear enough

Answer 7

the original problem is f(x) and it's asking to find f ' (x).

Answer 8

\[\Large f(x)\quad=\quad \int\limits\limits_{5-x^2}^{2013}\sqrt{10-t}\;dt\]Find \(\Large f'(x)\).

Ok now it makes more sense.

Answer 9

correct. sorry about that!

Answer 10

We'll need to apply the `Fundamental Theorem of Calculus, Part 1`:\[\Large \frac{d}{dx}\int\limits_a^x f(t)\;dt \quad=\quad f(x)\]

Answer 11

Before we do the actual math, try to think about what's happening a sec.
~We're Integrating (anti-differentiating)
~Then we're differentiating.

So we're doing something, then doing the opposite.

Answer 12

hold on. how do we know we don't use the fundamental theorem, part 2?

Answer 13

You can use the FTC Part 2 to do the integration, then after you've done that, apply the derivative operator.

But that whole process is summed up in the FCT Part 1.

Answer 14

FTC* blah

Answer 15

ah, okay. so if I start to apply theorem part 1, then would the first step be:

Answer 16

Ok think about this process a sec:\[\Large \frac{d}{dx}\int\limits\limits_a^x f(t)\;dt\quad=\quad \frac{d}{dx}F(t)|_a^x\quad=\quad \frac{d}{dx}\left[F(x)-F(a)\right]\]\[\Large =\quad f(x)-0\]F(a) is a constant, it's derivative is 0.

Answer 17

Where F(t) is the anti-derivative of f(t)

Answer 18

We started with some function, we integrate,
then the inside (arguments) of the function changes due to our boudaries,
then we differentiate, getting back what we started with,
but in terms of our new variable.

Answer 19

first, don't I need to reverse the bounds because the (5-x^2) should be on top?

Answer 20

Mmmm if you'd like to yes you can do that :)

Answer 21

You don't have to though, check this out a sec,

Answer 22

Let's try to apply this to our problem:\[\Large \frac{d}{dx}\int\limits\limits\limits_{5-x^2}^{2013}f(t)\;dt\quad=\quad \frac{d}{dx}F(t)|_{5-x^2}^{2013}\]
Let's not worry about what f(t) is right now, it's just the thing we're integrating.
It's integrates to give us `something` that we call F(t).
Evaluating at the boundaries gives us:\[\Large \frac{d}{dx}\left[F(2013)-F(5-x^2)\right]\]Then we take our derivative.

Answer 23

The fact that our x shows up in the lower boundary means that we'll have an extra negative showing up, right?
You would get the same result if you had just swapped the limits at the start anyway.

Answer 24

Confusing? :c

Answer 25

I'm trying to process everything you've said so far, sorry, I'm a slow reader x)

Answer 26

hehe :3

Answer 27

okay, so right now, I need to find the antiderivative of \[\sqrt{10-t} \] and then input in the boundaries? and then take the derivative of the result?

Answer 28

We DO NOT need to find it's antiderivative.
We are calling the antiderivative F(t), whatever that may be.
After plug in the bounds, we'll end up back at little f.

So we're not actually doing the integration of sqrt(10-t).

Answer 29

Here is a quick example:

Answer 30

GAH. Idk why this is going over my head. :/ sorry

Answer 31

\[\Large \frac{d}{dt}\int\limits \frac{e^{\sqrt{\arctan t^2}}}{t^2-4e^t}dt\]

Answer 32

So in the example here, the solution to this integral probably doesn't even exist.
But we're asked to integrate, and then afterwards, take the derivative (which will undo the integration).

So we can think of it like this:\[\Large \frac{d}{dt}\int\limits\limits \frac{e^{\sqrt{\arctan t^2}}}{t^2-4e^t}dt\quad=\quad \frac{d}{dt}\int\limits f(t)\;dt\]Where f(t) is that big messy thing.

Answer 33

"Integrating" would give us some function, we'll call F(t),\[\Large \frac{d}{dt}F(t)\]

Answer 34

Then we take it's derivative giving us:\[\Large f(t)\quad=\quad \frac{e^{\sqrt{\arctan t^2}}}{t^2-4e^t}\]

Answer 35

The thing we started with.

Does that make a little sense? :o

Answer 36

Or if we need to go even simpler:\[\Large \frac{d}{dt}\int\limits t^2\;dt \quad=\quad \frac{d}{dt}\frac{1}{3}t^3\]\[\Large =\quad t^2\]Yes..?

Answer 37

okay, I think I'm starting to see it. so if we go back to the original problem.

\[d/dx [F(2013) -F(5-x^2)]\]
=\[d/dx [\sqrt{10-2013} - \sqrt{10-5-x^2}]\]

Answer 38

\[\Large F(t)\quad\ne\quad \sqrt{10-t}\]It's the anti-derivative of that function.
We don't have a function to look at until we do the derivative :(

Answer 39

From here:\[\Large \frac{d}{dx}\left[F(2013)-F(5-x^2)\right]\]Taking the derivative of each term gives us:\[\Large \left[0-f(5-x^2)(5-x^2)'\right]\]
It gets a tad trickier lol
We have to apply the chain rule :(

Answer 40

if you're just applying the chain rule to (5-x^2) won't you just get -2x?

Answer 41

\[\Large \left[0-f(5-x^2)(-2x)\right]\]Yes good.
The function little f was the thing we started with (the square root).

Answer 42

\[\Large =2x\cdot f(\color{royalblue}{5-x^2})\quad=\quad 2x\cdot\sqrt{10-(\color{royalblue}{5-x^2})}\]

Answer 43

Where our original function was:\[\Large f(\color{royalblue}{t})\quad=\quad \sqrt{10-(\color{royalblue}{t})}\]

Answer 44

so essentially we now get back to what we started with except we have a new variable, (5-x^2), for t? we don't do anything with 2x?

Answer 45

correct.
We have a new argument, (5-x^2) in place of t,
and we get an extra factor of 2x outside due to the chain rule

Answer 46

=\[2x \sqrt{5+x^2}\]

Answer 47

so that is the final?

Answer 48

Mmmmmmm yah

Answer 49

Omg. Okay, thank you so much!! Sorry I was so slow x( Good thing I have 12 more problems just exactly like this, so it'll be good practice for me! Thanks again for being so patient, seriously!

Answer 50

ya hopefully it's a good start towards understanding the others D':

Answer 51

It is! I know the general idea and steps on what the heck I'm doing now, so I'm not so lost. :) Thx again