Question

Find integral of the following:

Answer 1

Hmm I think you posted this one the other day. Still having a bit of trouble with it?

We want to apply the FTC, Part 1. `Fundamental Theorem of Calculus`\[\color{brown}{\huge \frac{d}{dx}\int\limits_a^x f(t)dt=f(x)}\]

Answer 2

Want to see a simple example first? Because this one is really tough to get through if you don't understand the concept.

Answer 3

Yeah, the trig messes me up. That would be wonderful.

Answer 4

The idea is, we're taking an `integral` then we're undoing that integration by `differentiating`.

The thing that happens in between those steps is - The `argument` inside of our function changes due to what is located in our limits of integration.

So we should end up with what we started with, but it will be a function of a new variable now due to the boundaries.

Answer 5

\[\large \frac{d}{dx}\int\limits_1^x t^3 \; dt \qquad = \qquad \frac{d}{dx}\left[\frac{1}{4}t^4|_1^x\right]\]
Evaluating at the limits gives us,\[\large \frac{d}{dx}\left[\frac{1}{4}x^4-\frac{1}{4}1^4\right]\]Taking the derivative gives us,\[\large \left[x^3-0\right]\]See how we ended up with the same thing we started with? Except now it has an X as the variable instead of t.

Answer 6

Yeah, haha, wow. So you'll end up with the original problem huh?

Answer 7

Here's what is going to happen in our problem though... Let's pretend we had `more than just x` in the upper limit of integration.

Let's say we had this instead,\[\huge \frac{d}{dx}\int\limits\limits_1^{\color{orangered}{x^2}} t^3 \; dt\]

We'll have to do a little something extra, because the chain rule will come into play.

Answer 8

We'll end up with this,\[\large \frac{d}{dx}\left[\frac{1}{4}(x^2)^4-\frac{1}{4}(1)^4\right]\]
Taking the derivative will give us,\[\large (x^2)^3(x^2)'\]See how we have to apply the chain rule to the inside?

Answer 9

Yeah I see it. Never thought about using the CR.

Answer 10

Our answer is going to look really really really ugly.
But if you understand the concept, then it's not too bad.

Answer 11

\[\huge \frac{d}{dx}\int\limits\limits_{\sqrt3}^{\color{orangered}{x^2+\tan x^2}}\frac{7\pi \sqrt{t^4+t^2+1}}{\cos^2t+69}dt\]Woops lemme add alil color.

Answer 12

Yeah, I'm sure I can follow. Nothing is better than some guidance. thanks

Answer 13

In place of all the `t`'s, we'll put the orange stuff.
And then we'll apply the chain rule.

Answer 14

\[\frac{7\pi \sqrt{\color{orangered}{(x^2+\tan x^2)}^4+\color{orangered}{(x^2+\tan x^2)}^2+1}}{\cos^2\color{orangered}{(x^2+\tan x^2)}+69}\left(\color{orangered}{x^2+\tan x^2}\right)'\]

Answer 15

Scary huh? D':

Answer 16

Lol, very.

Answer 17

Then chain rule it?

Answer 18

Then we just take the derivative of the prime term and we're done!
I'm sure there is some simplification that could be done in there, but it's probably unnecessary.

Answer 19

There is one more `form` in which this problem shows up in that you'll want to be aware of.

Answer 20

Wait, the derivative of the prime term? What do you mean by that?

What would be the other form?

Answer 21

\[\huge \frac{d}{dx}\int\limits_x^3 \frac{1}{3}t^3\; dt\]See how the `lower` limit of integration has the variable term in it?
When we integrate, you'll see that we're actually `subtracting` the lower limit.
So a negative will pop into our answer.

Another way to remember it maybe is just to do this little trick.\[\large \int\limits_x^3 f(t)dt \quad = \quad -\int\limits_3^x f(t)dt\]

Answer 22

Oh sorry let's get back to the problem :)

Answer 23

In the big messy problem I pasted, I did the `setup` for the chain rule, I didn't actually take the derivative just yet.
The prime on the outside of the brackets is to show that we still need to differentiate that part.
It's the same as if there was a d\dx in front of it.

Answer 24

Ahh okay, I see. My professor always called it factoring it out. But that's just him...

Answer 25

Ah weird XD

Answer 26

Understand how to take the derivative of that last little part? :D

Answer 27

Umm

Answer 28

lol such a bum XD
Do you remember the derivative of tangent? :D

Answer 29

Ooops sorry I thought you meant something else. It'll be sec^2.

Answer 30

sec^2(x).

Answer 31

\[\large \left(x^2+\tan x^2\right)'\]
Yah sounds good, so we should get something like...\[\large \left(2x+2xsec^2x^2\right)\]Right?

Answer 32

Yeah take out the two and put it infront, right?

Answer 33

2(x+tan(x)+sec^2(x))

Answer 34

lol, did we forget the square on the x?

Answer 35

Why do you have a tangent in there...? :o

Answer 36

I simplified the trigfunction?

Answer 37

Or was it necessary?

Answer 38

I just don't understand what you did. Your derivative looks quite different than mine :O

Answer 39

Wait hold on, wouldn't we have to chain rule \[(\tan ^{2}(x))?\]

Answer 40

Was the upper boundary suppose to be this?\[\large x^2+\tan^2 x\]Because you have this on your homework sheet,\[\large x^2+\tan x^2\]

Answer 41

LOLOL oh shiz. you're right. My bad... you're right.

Answer 42

Which one is it suppose to be? XD hah

Answer 43

Mistook the square. :/

Answer 44

The second one

Answer 45

\[x ^{2}+tanx ^{2}\]

Answer 46

k cool c: Then check out my derivative that I posted above, I think that one is correct. tangent gave us secant sqared, then the x^2 chained to give us 2x.

Answer 47

Yeah, you're correct. Once that is done we do it to the rest?

Answer 48

rest of what? XD

Answer 49

Remember, we should end up with what we started with but with a small `chain rule` attached to it at the end.

We only needed to take the derivative of that last little part.

Answer 50

Hah, so that's it? We would just leave it at that?

Answer 51

\[\frac{7\pi \sqrt{\color{orangered}{(x^2+\tan x^2)}^4+\color{orangered}{(x^2+\tan x^2)}^2+1}}{\cos^2\color{orangered}{(x^2+\tan x^2)}+69}\left(\color{orangered}{2x+2x\sec^2 x^2}\right)\]

Yep! That's our solution!

Answer 52

Not at all what I expected, not at all... Got me by surprise. Do you think we could simplify? Or do you recommend leaving it at that?

Answer 53

Yah definitely leave it that way.
Under the square root, we have a 4th power, we really really really don't want to expand that out. It's way more trouble than it's worth.
And in the bottom, we're taking a trig function of a trig function.
We can't simplify that any more I'm afraid.

Answer 54

LOL, you're the expert here. I'll take your word. Thank you so much for taking the time to help me. :)

Answer 55

np c: