Question

Show that the function F of x equals the integral from 2 times x to 5 times x of 1 over t dt is constant on the interval (0, +∞).

Answer 1

https://gyazo.com/e5425cec35c6e331d016d9664296b104

Answer 2

@terenzreignz @zepdrix I used the fundamental theorem of calc to find derivative which was equal to 0.

Answer 3

That's enough to prove it right?

Answer 4

Use the second Fundamental Theorem instead.

Answer 5

Yes That's what was done :p

Answer 6

2nd, 1st it's all the same it's not a race ;)

Answer 7

Slope=0 means no change in function

Answer 8

It is not the same, the second part of the fundamental theorem for integral calculus is known also as Barrows rule, models the following equality:
\[\int\limits_{a}^{b}f(x)dx=F(b)-F(a)\]

Answer 9

I suppose I could start over, like I usually do :/
\[\Large \int\limits_{2x}^{5x}\frac1t dt \]

Answer 10

And yes, did you use the FTC as Owlcoffee stated it above ^
?

Answer 11

Yes

Answer 12

I used

Answer 13

Show me >:)

Answer 14

to be exact

Answer 15

\[F'(x)=u'(x)f(u(x))-v'(x)f(v(x))\]

Answer 16

\[\int\limits_{v(x)}^{u(x)}f(t)dt\]

Answer 17

It seems I may have to explain the second FTC after all :)

Answer 18

O_o why?

Answer 19

My teacher gave me that

Answer 20

Is it the way I wrote it? Cause I was too lazy to write out the complete thing

Answer 21

It has d/dx in front of the integral since that's what you need to do in this case

Answer 22

Well, *I* am giving this to you now:
Suppose you have a function F(x) such that the derivative of F is f on a given interval.
That is: F'(x) = f(x)

Then \[\Large \int\limits_a^b f(x) = F(a) - F(b)\]

Is the gist of it.

Answer 23

Find the derivative.

Answer 24

Ok

Answer 25

Things don't seem to be going my way ^^
\[\Large \int\limits_a^b f(x)dx = F(b) - F(a)\]

Answer 26

There, use that.
Now, can you integrate 1/t?

Answer 27

ln(5x)-ln(2x)?

Answer 28

That's right ^^
Simplify it using the rules concerning logarithms:
I'd tell you which rule exactly, but maybe you already know :D

Answer 29

A refresher would be nice

Answer 30

Wait

Answer 31

Just remembered

Answer 32

\[\ln \frac{ 5x }{ 2x }\]?

Answer 33

Yup. Simplify further?

Answer 34

ln (5/2)

Answer 35

Which is constant.
Et voila ^^

Answer 36

So is what I did considered wrong?

Answer 37

I'm not exactly sure what you did :>
But you do have that tendency to overcomplicate things sometimes :D
But that which you used looked nothing like what was needed in this particular question... unless I'm mistaken.

Answer 38

My teacher complicates it not me :p I'm using her formula.

Answer 39

Well, that's why I'm here :D

Answer 40

https://gyazo.com/cb41c86a344c6fb6dbadfadabd886845

Answer 41

That's what I was given.

Answer 42

I used a variant of the 1st which may seem complicated but it was plug and play as well.

Answer 43

Part II comes to mind...

Answer 44

It seems simpler but I don't feel like erasing and rewriting

Answer 45

So I think I'm good with what I wrote

Answer 46

:)