Question

Evaluate:

Answer 1

\[(d/dx) \int\limits_{x^2}^{10} (dz / (z^2 + 1))\]

Answer 2

Show steps please

Answer 3

Well first things first, do you know the antiderivative for the integral?

Answer 4

Wait lol nvm

Answer 5

didnt see that part in front.

Answer 6

\[\Large \int_{x^2}^{10} \frac{dz}{z^2+1}=F(10)-F(x^2) \]
Can you tell from here what happens if you differentiate?

Answer 7

Ah nvm spacelimbus, you win!

Answer 8

Yeah that makes sense to me the F(10) - F(x^2)

Answer 9

*smiles* I have had my troubles with these kind of exercises @vf321, hence I had to LaTeX this one out *laughs*

Answer 10

Differentiate the RHS @ConfusedAboutCalculus, that's what you are looking for.

Answer 11

So I get the derivative of F(10) - F(x^2)? How?

Answer 12

Well the first one is a number correct? Any Antiderivative evaluated at a set value: \(F(10)\), if you differentiate you a number, what do you get?

Answer 13

zero

Answer 14

exactly, the second one is a bit more tricky, because it's an Antiderivative (which is easy to differentiate) but inside there is another function y=x^2, so you need to apply which rule to differentiate that?

Answer 15

2x, power rule

Answer 16

true, but all in all that's the chain rule right?

Answer 17

Yup!

Answer 18

good, so do that and you're done

Answer 19

but what about the dz and z^2 + 1 and all that

Answer 20

\[ \Large F(10)-F(x^2) \]
Differentiate:

\[\Large 0 - f(x^2)2x \]

Answer 21

If your Antiderivative is a function of x, and you differentiate that again, your derivative will be a function of x too.

Answer 22

I don't understand what you mean. What happens to the Z variables

Answer 23

Think about an easy example

\[ \Large \int_0^x zdz =\frac{x^2}{2} \] If you differentiate the righthandside you get \(x\) which is of the same order than your standard function, just with a different variable.

Answer 24

in other words, you don't have to worry about the original function anymore it will be the same just with a different substitution for its variables. (Plus the chain rule in this case)

Answer 25

makes any sense @ConfusedAboutCalculus ?

Answer 26

Not really.. haha. Since it's the integral from x to 10 of (dz / (z^2 + 1)), I thought you first had to reverse the chain rule and then do F(10) - F(x^2)

Answer 27

and I thought that f(x) was dz / (z^2 + 1)

Answer 28

\[ \Large \int_{x^2}^{10} \frac{dz}{z^2+1}=F(10)-F(x^2)\]
Do you agree with me that when we differentiate this we get the following (applying Leibniz differential operator on both sides)

\[ \large \frac{d}{dx}\int_{x^2}^{10} \frac{dz}{z^2+1}=\frac{d}{dx}(F(10)-F(x^2))\]

The LHS is too hard to evaluate directly, fortunately by the fundamental theorem of Calculus it is equal to the RHS of the equation, so lets differentiate that instead.

\[\Large \frac{d}{dx}(F(10)- \frac{d}{dx}F(x^2) \]

Answer 29

so I I thought that if I took the anti derivative of dz / (z^2 + 1), then do F(10) - F(x^2), then find the derivative of that result

Answer 30

Now for the second one you have to apply the chain rule, derivative of the outside (with respect to the inside) times the derivative of the inside function (inside function is y=x^2, that's the lower bound of your integral)

Answer 31

So I understand it equals 0 - f(x^2) (2x)

Answer 32

\[ \Large \frac{d}{dx}F(x^2)=F'(x^2)2x \]
and F'(anything)=f(anything)

Answer 33

Is that it?

Answer 34

-f(x^2)(2x) ?

Answer 35

well the result is \[ \Large f(x^2)2x \]

And you said already yourself that

\[ \Large f(z)= \frac{1}{1+z^2} \]

So \[ \Large f(x^2)2x= \frac{2x}{1+x^4}\]

Answer 36

Back-Substitution.

Answer 37

why isn't it negative? if it's f(10)(0) - f(x^2)(2x), shouldn't it be -f(x^2)(2x)? then you substitute that?

Answer 38

of you're right there, it's negative. I did back-substitution but the results wants it negative, as derived above.

Answer 39

\[ \large 0 - f(x^2)2x=- \frac{2x}{1+x^4}\]

Answer 40

because x^2 was the lower bound, not the upper bound.

Answer 41

I think there is an easier way to do this

Answer 42

If you discovered that, please let me know (-:

Answer 43

so what If I first did the integral from x^2 to 10 of 1/ (z^2 + 1) and then I got

tan^-1(10) - tan^-1(x^2). Then I take the derivative of that and get:

0 - (1/ 1+ x^4)(2x)

Answer 44

\[\Large \int_{x^2}^0 \frac{1}{z^2+1}dz= \tan^{-1}(z) = - \tan^{-1}(x^2)\]

Answer 45

final answer is -2x / (1+x^4)

Answer 46

Very correct, and you know why this works?

Answer 47

yeah because I used the fundamental theorem of calculus and then took the derivative

Answer 48

the derivative is over the whole integral

Answer 49

Simply put, the integral is easy as hell.

Try the following

\[\Large \frac{d}{dx}\int_0^{x^5} \frac{e^{2z}}{1+z^6}dz \]

Answer 50

There is no easy integral method for this, but you can still apply the fundamental theorem of Calculus (-:

Answer 51

I exaggerated a bit with this example, you might want to look for a better one, but as soon as there is an integral you can't easily integrate, you might want to use the FTC instead, over trying to integrate the function manually and then differentiate.

Answer 52

so the FTC is just a shortcut for integration?

Answer 53

Well, lets say if you want to discuss the limits of an integral and so on, it's critical points than the FTC of important. Otherwise what you do is very correct and rigorous, but it can't always be applied.

Answer 54

I see well thank you for your help I learned a lot from the other method you showed me

Answer 55

You're welcome.

Answer 56

\[\text{Nice}\]