Question

find F'x . F (x)= (integral sign) bounds [1,x] t^(1/4)dt.

Answer 1

\[F(x) = \int\limits_{1}^{x}t^{1/4}dt\]

Answer 2

yupp

Answer 3

For any equation, by the fundamental theorem of calculus, I get:
\[\int\limits_{a}^{b}f'(x) = f(b)-f(a)\]

In other words, the result of the integral of t^(1/4) is the function whose derivative is t^(1/4)

Answer 4

so...1/4t^-3/4 is the derivative

Answer 5

Nono...
think of it this way:
If you were going to DO that integral, your goal would be to find the antiderivative. You're trying to find something that you could derive and get t^(1/4)
right?

Answer 6

thn tht is \[4/5t ^{5/4}\]

Answer 7

Right, but that's a little extra work, it turns out. Because the next thing they ask you to do is to take the derivative of that...
And you just put a lot of work into handcrafting that thing specifically so that it's derivative would end up being t^(1/4)

Answer 8

ohh..so its just \[x ^{1/4}\]

Answer 9

Yes =)

Answer 10

Another way to think of it is just that integrating and deriving are opposites, so they undo each other.

Answer 11

n oh! btw, to find F'x for F(x)-\[\int\limits_{-x}^{x}t ^{3}= 0\]?

Answer 12

why does it equal 0

Answer 13

Is that t^3?

Answer 14

yup

Answer 15

Oh, will in this case, it's a definite integral. So I take the antiderivative and evaluate at the limits...
it just becomes one constant minus another constant.
The derivative of a constant is just 0.
Notice that in the last problem, the limits of the integral involved a variable. So when I evaluate the integral at the limits, there's a variable still involved.

Answer 16

\[t ^{4}/4\] at -x its 1/4 and at x its 1/4...so 1/4-1/4= 0...?

Answer 17

Oh, those aren't constants. The first time I read it, I thought that was from -pi to pi.
Okay, you get (1/4)t^4 yes
then evaluate from -x to x:
(1/4)x^4 - (1/4)(-x)^4
x^4 = (-x)^4, so that gives 0.

Answer 18

the answer is x^(4/5)