OpenStudy (anonymous):

Find integral of the following:

5 years ago
OpenStudy (anonymous):
5 years ago

zepdrix (zepdrix):

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`\[\color{brown}{\huge \frac{d}{dx}\int\limits_a^x f(t)dt=f(x)}\]

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

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.

5 years ago
zepdrix (zepdrix):

\[\large \frac{d}{dx}\int\limits_1^x t^3 \; dt \qquad = \qquad \frac{d}{dx}\left[\frac{1}{4}t^4|_1^x\right]\] 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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

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.

5 years ago
zepdrix (zepdrix):

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?

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

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

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

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

5 years ago
zepdrix (zepdrix):

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

5 years ago
zepdrix (zepdrix):

Scary huh? D':

5 years ago
OpenStudy (anonymous):

Lol, very.

5 years ago
OpenStudy (anonymous):

Then chain rule it?

5 years ago
zepdrix (zepdrix):

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.

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

Wait, the derivative of the prime term? What do you mean by that? What would be the other form?

5 years ago
zepdrix (zepdrix):

\[\huge \frac{d}{dx}\int\limits_x^3 \frac{1}{3}t^3\; dt\]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.\[\large \int\limits_x^3 f(t)dt \quad = \quad -\int\limits_3^x f(t)dt\]

5 years ago
zepdrix (zepdrix):

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

5 years ago
zepdrix (zepdrix):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

Ah weird XD

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

Umm

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

sec^2(x).

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

lol, did we forget the square on the x?

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

I simplified the trigfunction?

5 years ago
OpenStudy (anonymous):

Or was it necessary?

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

Which one is it suppose to be? XD hah

5 years ago
OpenStudy (anonymous):

Mistook the square. :/

5 years ago
OpenStudy (anonymous):

The second one

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

rest of what? XD

5 years ago
zepdrix (zepdrix):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

\[\frac{7\pi \sqrt{\color{orangered}{(x^2+\tan x^2)}^4+\color{orangered}{(x^2+\tan x^2)}^2+1}}{\cos^2\color{orangered}{(x^2+\tan x^2)}+69}\left(\color{orangered}{2x+2x\sec^2 x^2}\right)\] Yep! That's our solution!

5 years ago
OpenStudy (anonymous):

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?

5 years ago
zepdrix (zepdrix):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

np c:

5 years ago