Question

I need help with an intergration quetions, see the attached photo.

Answer 1

so what is the fundamental theorem of calc?

Answer 2

also, logically, how do you undo taking a derivative?

Answer 3

by FTC will it be -3x^6+2^6Sqrt(2) + coz(2^2-9)

@FibonacciChick666

Answer 4

Hmm, your first `term` looks correct.
But you plugged a bunch of 2's into the rest of it... confusing.

Answer 5

@zepdrix my mistake, it should be -3x^6+x^4sqrt(x)+ cos(x^2-9) ?

Answer 6

\[\large\rm \frac{d}{dx}\int\limits_c^x f(z)dz=f(x)\]

\[\rm \frac{d}{dx}\int\limits\limits_c^x -3\color{orangered}{z}^6+\color{orangered}{z}^4\sqrt{\color{orangered}{z}}+\cos(\color{orangered}{z}^2-9)~dz\quad=\quad-3\color{orangered}{z}^6+\color{orangered}{x}^4\sqrt{\color{orangered}{x}}+\cos(\color{orangered}{x}^2-9)\]Ah yes, much better! :)

Answer 7

what about if the upper bound is not a x?

Answer 8

You'll run into a bunch of different variations of this type of problem,
so that's a good question to be asking!

Answer 9

for instance this this question

Answer 10

\[\large\rm \frac{d}{dx}\int\limits_{2\sqrt2}^{2x^4}\cos(t^2+3)~dt\]Remember that integration is `anti-differentiation`.
So try to think about what is happening in this process.
You're anti-differentiating,
then you're differentiating,
getting back the thing you started with.

What happens between those steps,
is that the argument of the function changes, ya?

Answer 11

Let's generalize this a lil bit, instead of some ugly function cos(stuff),
let's call it f(t)\[\large\rm \frac{d}{dx}\int\limits\limits_{2\sqrt2}^{2x^4}f(t)~dt\]We're going to some "pseudo-integration" here.
We're not actually looking for an anti-derivative,
we'll just assume that one exists, and call it F(t).
Ok, so integrating gives us,\[\large\rm \frac{d}{dx}\int\limits\limits\limits_{2\sqrt2}^{2x^4}f(t)~dt\quad=\quad\frac{d}{dx}F(t)|_{2\sqrt2}^{2x^4}\]And then we have to evaluate this at our bounds,\[\large\rm \frac{d}{dx}\left[F(2x^4)-F(2\sqrt2)\right]\]And we can FINALLY take our derivative,\[\large\rm =f(2x^4)(2x^4)'-0\]F turns back into f,
but we have to apply our chain rule!

The other term is just a constant,
so it's derivative is zero.

Answer 12

TL DR: Chain rule when you have `more than just x as a boundary`

Answer 13

\[\large\rm =f(2x^4)(8x^3)\]Recall our function we started with:\[\large\rm f(\color{orangered}{t})=\cos(\color{orangered}{t}^2+3)\]Because plugging your bound into the final answer can be tricky.

Answer 14

Thanks for covering zep! This whole no notifiertions thing is quite annoying

Answer 15

ye :c

Answer 16

is this correct
f'(x) = cos(x^2+3).(8x^3) ?

Answer 17

Close! :)
Our upper bound is not x though, so we're not plugging x into the function f.
We're plugging more than that in.\[\large\rm f(\color{orangered}{t})=\cos(\color{orangered}{t}^2+3)\]\[\large\rm f(\color{orangered}{2x^2})=\cos(\color{orangered}{(2x^2)}^2+3)\]And keep in mind, this is NOT f'(x).
This is just f(x), the function we started with, but with a new argument/variable inside of it.

Answer 18

@zepdrix

you lost me a little by evaluation the function at the upper and lower bounds, because I am used to the methods of using the 1st FTC. hence replacing the upper bound in this question with U = 2x^4

Answer 19

Ah yes, I guess I was using the FTC Part 2 quite a bit in there, hmm.

Answer 20

hold up, I think i spotted my error. let me evaluate again then you can check ?

Answer 21

I guess the general formula was too general.
Let me say it this way:\[\large\rm \frac{d}{dx}\int\limits_c^{g(x)}f(t)dt\quad=\quad f(g(x))\cdot g'(x)\]Maybe that's more confusing though.

Answer 22

k :)

Answer 23

If you get the time,
here is another problem to practice:\[\large\rm \frac{d}{dx}\int\limits\limits_{2x^4}^{2\sqrt2}\cos(t^2+3)~dt\]

Answer 24

so \[dy/dx = \cos(2x^6 + 3).8x^3\]

Answer 25

Oh oh oh :( I called it 2x^2 earlier, my bad.

Answer 26

\[\large\rm f(\color{orangered}{2x^4})=\cos(\color{orangered}{(2x^4)}^2+3)\]We're plugging in 2x^4,
I think you have the wrong exponent rule: \(\large\rm (x^4)^2\ne x^6\)

Answer 27

And don't forget to square the 2 as well.

Answer 28

ohhh ok
\[f(2x^4) = \cos((4x^8)+3).8x^3\]

Answer 29

Yay good job \c:/

Answer 30

Uh well, maybe not call it f,
call it y' like you were doing before.

Answer 31

Because f was only the cosine portion.

Answer 32

Ok thank you and sorry for my erros

and your practice question
\[\frac{ d }{ dx }\int\limits_{2x4}^{2\sqrt{2}}\cos(t^2+3)dt = -\frac{ d }{ dx }\int\limits_{2\sqrt{2}}^{2x^4}\cos(t^2+3)dt \]

then just do the same process?

Answer 33

Good good good :)
How bout this one?\[\large\rm \frac{d}{dx}\int\limits\limits_{2x^4}^{x^2}\cos(t^2+3)~dt\]Hint: Apply another integral rule.\[\large\rm \int\limits_a^c f(t)dt\qquad=\qquad \int\limits_a^b f(t)dt+\int\limits_b^c f(t)dt\]Where \(\large\rm a<b<c\).

Answer 34

I have not done this type of problems but using your hint I would my attempt would be
\[\frac{ d }{ dx }\int\limits_{2x^4}^{x^2}\cos(t^2+3)dt = \frac{ d }{ dx }\int\limits_{0}^{2x^4}\cos(t^2+3)dt - \frac{ d }{ dx }\int\limits_{0}^{x^2}\cos(t^2+3)dt \]

Answer 35

Ah very close! :)\[\large\rm \frac{d}{dx}\int\limits_{2x^4}^{x^2}f(t)~dt\quad=\quad \frac{d}{dx}\int\limits_{2x^4}^{b}f(t)~dt+\frac{d}{dx}\int\limits_{b}^{x^2}f(t)~dt\]Where b is some value between them.
Then,\[\large\rm \frac{d}{dx}\int\limits\limits_{2x^4}^{x^2}f(t)~dt\quad=\quad -\frac{d}{dx}\int\limits\limits_{b}^{2x^4}f(t)~dt+\frac{d}{dx}\int\limits\limits_{b}^{x^2}f(t)~dt\]It looks like the negative should go on the first one though, ya?

Answer 36

These are the three or four main forms that this type of problem will show up in.
So these are nice examples to at least consider :)

Answer 37

yes, but if you are not told that b is a value between the upper and lower bound, will my methods still work?

Answer 38

The fact that you picked 0 to go between them is fine.
Sometimes that will cause a problem, but not usually.

I wasn't being picky about that.
I was stressing the fact that you need to apply your negative rule to the FIRST one.
So you have the negative on the wrong one.

If you're going some distance `from a` `to c`,
that's the same as going `from a` `to b` then going `from b` `to c`.

We're taking the bounds of our integral, and breaking it up into two ... paths,
if that makes sense.

But what you wrote is traveling `from b` `to a` then going `from b` `to c`

Answer 39

ok I understand now. Thank you.

Answer 40

Cool, yay team!\[\large ୧ʕ•̀ᴥ•́ʔ୨\]

Answer 41

Thank you @zepdrix @FibonacciChick666
I appreciate your time, I have given the best response and followed.

All the best! :)

Answer 42

@zepdrix good luck saving the maths world, from the villains of math problems.

Answer 43

XD