Question

http://prntscr.com/9rcl4u

Answer 1

@robby8974

Answer 2

I would recommend rewriting the inside as t^-4 just for simplicity

Answer 3

−1/3x^6+1/192

Answer 4

But you should explain where that comes from

Answer 5

all you do is integrate using the power rule

Answer 6

But she may not know that so it's best to explain with words what you did

Answer 7

i know you integrate. so because it's asking for F'(x), you don't find the original equation of F(x) because that is the derivative, am i correct?

Answer 8

if you put it into www.mathway.com you will find out that i am right (i used it to check my work)

Answer 9

and yes you are right

Answer 10

but first what are the answer choices

Answer 11

there is no answer choices

Answer 12

you're just plugging in x^2 and 4 into t right?

Answer 13

yes

Answer 14

Yes Amy, this is an application of the Fundamental Theorem of Calculus, Part 1

Since your upper limit is more than just x,
you'll have to chain rule.

So you'll get a lil bit more than just plugging x in :)

Answer 15

do the chain rule for (x^2)^4?

Answer 16

yes

Answer 17

\[\large\rm F(x)=\int\limits_c^x g(t)dt\]Then FTC, Part 1 tells us:\[\large\rm F'(x)=g(x)\]You're integrating, then differentiating.
These two things undo one another.
All that changes is the argument of the function.
So instead of t, you have x for the argument of your function.

But when we have something like this:\[\large\rm F(x)=\int\limits_c^{x^2}g(t)dt\]Now, when we take our derivative, we'll have to chain rule because the inner function is x^2.\[\large\rm F'(x)=g(x^2)\cdot (x^2)'\]

Answer 18

So no, not chain rule for that entire thing, only the x^2, the inner function.

Answer 19

Confused? :o
Too much?

Answer 20

so take the chain rule like in differential cal?
like y'=x^2=2x

Answer 21

Yes, here is the long way using FTC, Part 2,
maybe it will help to see what is going on.
\[\large\rm F(x)=\int\limits_4^{x^2} g(t)dt\]Let's call the anti-derivative of g, G for simplicity.
Integrating gives us,\[\large\rm F(x)=G(t)|_4^{x^2}\]Evaluating at our limits gives us,\[\large\rm F(x)=G(x^2)-G(4)\]Now we differentiate.
Realize that G is going to turn back into little g.
But G(4) is just a constant! Will become zero in this process.\[\large\rm F'(x)=g(x^2)(x^2)'-0\]\[\large\rm F'(x)=2x~g(x^2)\]

Answer 22

To answer your question, yes.
Sorry if I made that over-complicated :) lol

Answer 23

ok makes sense, so I got 8x^7

Answer 24

alright

Answer 25

Well in this problem: \(\large\rm g(\color{orangered}{t})=\dfrac{1}{\color{orangered}{t}^4}\)

So if we end up with:\[\large\rm F'(x)=2x~g(\color{orangered}{x^2})\]I guess we should get something like:\[\large\rm F'(x)=2x\frac{1}{(\color{orangered}{x^2})^4}\]Right?

Answer 26

yes

Answer 27

Which is not 8x^7
+_+

Answer 28

doesn't (x^2)^4=8x^7?

Answer 29

(x^2)^4 = x^8

You don't need to take a derivative again.

Answer 30

oh you're just multiplying it?