Find integral of the following:

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

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

6 years agoYeah, the trig messes me up. That would be wonderful.

6 years agoThe 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.

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

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

6 years agoHere'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.

6 years agoWe'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?

6 years agoYeah I see it. Never thought about using the CR.

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

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

6 years agoYeah, I'm sure I can follow. Nothing is better than some guidance. thanks

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

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

6 years agoScary huh? D':

6 years agoLol, very.

6 years agoThen chain rule it?

6 years agoThen 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.

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

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

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

6 years agoOh sorry let's get back to the problem :)

6 years agoIn 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.

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

6 years agoAh weird XD

6 years agoUnderstand how to take the derivative of that last little part? :D

6 years agoUmm

6 years agolol such a bum XD Do you remember the derivative of tangent? :D

6 years agoOoops sorry I thought you meant something else. It'll be sec^2.

6 years agosec^2(x).

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

6 years agoYeah take out the two and put it infront, right?

6 years ago2(x+tan(x)+sec^2(x))

6 years agolol, did we forget the square on the x?

6 years agoWhy do you have a tangent in there...? :o

6 years agoI simplified the trigfunction?

6 years agoOr was it necessary?

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

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

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

6 years agoLOLOL oh shiz. you're right. My bad... you're right.

6 years agoWhich one is it suppose to be? XD hah

6 years agoMistook the square. :/

6 years agoThe second one

6 years ago\[x ^{2}+tanx ^{2}\]

6 years agok 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.

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

6 years agorest of what? XD

6 years agoRemember, 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.

6 years agoHah, so that's it? We would just leave it at that?

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

6 years agoNot 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?

6 years agoYah 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.

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

6 years agonp c:

6 years ago