Question

If f(x) = integral of 1/ (sqrt of t^3 +2)dt, then f'1 = ?
I'll write the equation with teh equation thing also

Answer 1

is the 2 under the radical sign?

Answer 2

\[If f(x) = \int\limits_{0}^{2x} 1/ (\sqrt{t^3+2}) dt, \]
then f' (1) = ?

Answer 3

oh it's a fundamental theorem of calculus problem right?

Answer 4

yeah

Answer 5

the upper limit says 2x btw

Answer 6

yes

Answer 7

help me!

Answer 8

http://openstudy.com/study#/updates/4f0a05e7e4b014c09e644eea

Answer 9

Well this problem is gonna bug me if I don't figure it out, so give me a sec...

Answer 10

r u still there

Answer 11

yes, but I'm a bit thrown by how they want you to get the value of the integral...
perhaps I'm just out of practice on these :/

Answer 12

=( so u dont know how 2 approach teh problem?

Answer 13

ok the derivative is the integrand, but you need the chain rule here

Answer 14

\[ f(x) = \int\limits_{0}^{2x} \frac{1}{\sqrt{t^3+2}}dt,\]
\[f'(x)=\frac{1}{\sqrt{(2x)^3+2}}\times 2\] via the chain rule.
clean this up with some algebra or just replace x by 1 to get
\[f'(1)\]

Answer 15

oh right, so the f(0) doesn't matter then...

Answer 16

i get
\[f'(1)=\frac{2}{\sqrt{10}}\]

Answer 17

f'(0) I mean*

Answer 18

\[f(0)=0\]

Answer 19

?
how do you know?

Answer 20

the lower limit of integration gives you the constant. the only work here to to write the derivative making sure to use the chain rule because the upper limit is 2x and not x

Answer 21

oh right XD

Answer 22

\[f(0)=\int_0^0f(t)dt=0\]

Answer 23

so it's not that f(0) is zero by itself, it disappears because we took the derivative
and the derivative of a constant is zero

Answer 24

\[f(x)=\int_a^xf(t)dt\] is a function of x. if you replace x by a you get
\[f(a)=\int_a^af(t)dt=0\]

Answer 25

ok I see what you're saying...

Answer 26

that is how we know that for example
\[\int_1^2t^3dt=\frac{2^4}{5}-\frac{1}{5}\] because we know that both
\[f(x)=\int_1^2t^3dt\] and
\[g(x)=\frac{x^3}{4}+C\] have the same derivative, and we also know that
\[f(1)=0\] meaning that
\[g(1)=0=\frac{1}{4}+C\]

Answer 27

typo up top, denominators should be 4!

Answer 28

f(x) can't have constant bounds on both ends though can it? that makes it a constant, 7/2, the value of the integral

Answer 29

so the derivative of f(x) the way you wrote it would be zero ...no?
sorry to carry on in your post sabrina

Answer 30

not sure exactly what you mean
\[\int_a^b f(t)dt\] is a number, whereas
\[\int_a^xf(t)dt\] is a function of x

Answer 31

and the derivative of
\[\int_a^xf(t)dt\] is
\[f(x)\] by ft of c

Answer 32

no problem..u asking questions calrifies many things for me also

Answer 33

maybe i can explain this way, see if it makes sense. with a simple example. suppose i write
\[f(x)=\int_1^x2tdt\] this is a function of x. and clearly
\[f(1)=\int_1^12tdt=0\] now suppose i would like to find
\[f(3)=\int_1^32tdt\] i know two functions whose derivative is
\[2x\] one is
\[f(x)=\int_1^x2tdt\] and the other is
\[x^2\]

Answer 34

sat when u are free, please see my last question

Answer 35

because i have two functions with the same derivative, i know they can at most differ by a constant. and what is that constant? well since
\[f(1)=0\] i know the constant must be
\[-1\] since if i evaluate
\[x^2\] at 1 i get 1 and thefore i know that
\[f(x)=\int_1^x2tdt=x^2-1\]

Answer 36

this allows me to evaluate
\[f(x)\] at any number i choose. so
\[f(3)=\int_1^32tdt=3^2-1\]and so on

Answer 37

but you wrote above\[f(x)=\int_{1}^{2}t^3dt\]which is a constant
I think it was just a typo, I understand the rest of what you are saying
thanks I really do get it better now.