Question

Please help! Will FAN AND MEDAL!

Find the area under the graph of f(x) = e-lnx(x) on the interval [1, 2].

a. 0.5
b. -0.5
c. 0.693
d. -0.693

Answer 1

Can you please help? @zepdrix @ganeshie8 @Directrix

Answer 2

\[area = \int\limits_{a}^{b} f(x) dx\] http://prnt.sc/axbnia

Answer 3

take the anti derivative of the function and then find F(b)-F(a)
where F represents the(anti derivative )

Answer 4

Okay so will I replace x with 1 and 2?

Answer 5

\[f(x)=e -\ln(x)\] this is the equation right

Answer 6

yeah that's what I want to know too lol

Answer 7

i think that is a typo

Answer 8

or

\[e^{lnx} ?\]

Answer 9

find take the anti derivative and then yes substitute x for 2 and 1

Answer 10

hm

Answer 11

\[e^{-\ln(x)}=e^{\ln(\frac{1}{x})}=...\]

Answer 12

\[e^-{lnx(x)}\]

Answer 13

\[e^{-\ln(x)} = e^{\ln(\frac{ 1 }{ x })} = \frac{ 1 }{ x } = ? \]

Answer 14

e^-lnx(x) this is the equation.

Answer 15

I

Answer 16

essentially that equals 1/x

Answer 17

I am truly confused. Can someone please show me in steps on how to solve this?

Answer 18

\(\large\rm e^{-(ln x)x}\)
This?

Answer 19

\(\large\rm x e^{-(ln x)}\)
Or this maybe?

Answer 20

I think this is what the question was

\[e^{-\ln(x)} = e^{\ln\frac{ 1 }{ x }} = \frac{ 1 }{ x }\]

\[\int\limits_{1}^{2} \frac{ dx }{ x }\]

Answer 21

Aryana, you need to make your question more clear...

Answer 22

The equation is \[e ^{-lnx(x)}\]

Answer 23

you can actually just re-write that as 1/x

Answer 24

Ohh okay then I guess that is what my new equation would be.

Answer 25

Both x's are in the exponent like that?
Ok.

Answer 26

This is what you're integrating

\[\int\limits_{2}^{1} \frac{ dx }{ x }\]

Answer 27

sorry, bounds of integration are wrong lol 2 should be on top.

Answer 28

\[\large\rm e^{-\ln(x)\cdot x}\quad=e^{\ln(x^{-1})\cdot x}\quad=e^{\ln(x^{-x})}=x^{-x}\]

Answer 29

It's certainly not 1/x that you're integrating.
Not if you have 2's x's up there like that.

Answer 30

2* x's*

Answer 31

Yes the x's are right next to each other like x(x). That is what was confusing me.

Answer 32

what we need is a screen shot perhaps
you can't really have two x's next to each other
i mean you can, but you shouldn't

Answer 33

One second I will send a screen shot.

Answer 34

@satellite73 @zepdrix @greatlife44 @Nnesha

Answer 35

Personally, I think it's a typo,
because if it had been \(\large\rm e^{-ln x}\),
then it would actually correspond to one of your options,
as greatlife was indicating.

Answer 36

so actually if you were to just integrate this

Answer 37

http://prnt.sc/axbvxg

Answer 38

\[\int\limits_{1}^{2} \frac{ dx }{ x } = \ln|x|+C\]

Answer 39

I got 0.5 but I don't know if that is correct. I think I might have input it wrong.

Answer 40

@greatlife44 @Nnesha @zepdrix

Answer 41

so first the integral of 1/x is ln|x|+C

now we evaluate it

Answer 42

\[\ln|x|+C \] from 2 to 1 =\[Ln(2)-Ln(1)\]

Answer 43

ln(1) = 0 so you're left with Ln(2).

just exactly how did you get 0.5?

Answer 44

Ohh okay. I see how you got Ln(2) which equals .0693

Answer 45

it really says \[e^{-x\ln(x)}=x^{-x}\] i have no idea to how to that