Calculate this and get a FullInfoAccount

[Colster]

the problem is you have to take the integral from the f(x) intersection to 1.

A(square) = 1*1 = 1
A(square)/2 = 0.5
x interception = z

integral{x0to[1]}(f(x)) = 0.5
[x*ln(x)-x+a*x] {x0to[1]} = 0,5


1*ln(1)-1+a*1 - [x0*ln(x0)-x0+a*x0] = 0.5 --> equation 1

ln(x0) + a = 0
ln(x0) = -a
x0 = e^(-a)

x0 -> equation 1

-1 + a - (e^(-a))*ln(e^(-a)) + e^(-a) - a*e^(-a) = 0,5 |+1
a + a*e^(-a)) + e^(-a) - a*e^(-a) = 1,5
a + ( e^-a) = 1,5

solution 1: 1.198290437
solution 2: -.8576766739

2 doesnt make any sense, so its

test:

f(x) = ln(x) + 1.198290437
x0 = e^(-1.198290437) = .3017095628

integral from .3017095628 to 1 f(x) = 0.5

problem solved, its 1.198290437

edit: so camicio is right, just wanted to solve the problem, i like mathematics (flames inc )

[Ahskrew]

Ah, i now see what i did wrong, haha. ty Colster

[blubb12345]

a = 1.2344720351728634183955884763908

okey, edited now, my guess is on this one

WAA YOU ARE THE WINN0R

1.2344720351728634183955884763908

[guitargod218]

The answer is definitely over 9000 (never mind, it was answered already hehe).

I'm a high school sophomore, and I'm very interested in how this problem is solved. However, I don't think I've ever been taught how to solve this (I'm in the equivalent of AP Algebra II this year and am going into AP Pre-calculus next year). Is this a calculus problem?

[wowpew]

What was wrong with my answer: a = -ln(x)+x ?

[Dendra]

Ah I didn't see this...I could have done it lol.
The maths isn't that hard, it just requires a bit of thinking to work out what equations to solve

[ReidE96]

wrong :P (integral of ln(X) is x*ln(x)-x (x>0))

|||||||||
vvvvvvvv
Integral of f(x) = integral of ln(x) + ax +c
=xlnx + ax + c - integral of 1
=xlnx +(a-1)x + c

i am sorry, its wrong
and the rest is also wrong

Actually, no, that bit isn't wrong. f(x) = ln(x) + a.
integral ( f(x) )
= integral ( ln(x) + a )
= integral ( ln(x) ) + ax + c

Using integral uv' = uv - integral vu',
integral ln(x)
= xln(x) - integral 1
= xln(x) - x

Hence,
integral ( ln(x) ) + ax + c
= xln(x) - x + ax + c
= xln(x) + (a-1)x + c

And it's what I said! Whay!

(yes, I know it's been solved)

[camicio]

Lol, you're right blubb12345. Do more giveaways like this please, I'll solve the next one hopefully .
guitargod218 you just need to know how to integrate logarithms, know what an integral is and avoid doing stupid errors or making stupid assumptions.
tamen, a is supposed to be a real constant number.

[Ahskrew]

Think this one was the hardes math problem ive ever solved tbh, haha and i got a account out of it sweet and ty

[blubb12345]

Actually, no, that bit isn't wrong. f(x) = ln(x) + a.
integral ( f(x) )
= integral ( ln(x) + a )
= integral ( ln(x) ) + ax + c

Using integral uv' = uv - integral vu',
integral ln(x)
= xln(x) - integral 1
= xln(x) - x

Hence,
integral ( ln(x) ) + ax + c
= xln(x) - x + ax + c
= xln(x) + (a-1)x + c

And it's what I said! Whay!

(yes, I know it's been solved)

uh sry missed the

=xlnx + ax + c - integral of 1

saw that the -x missed



your mistake is that you chose the wrong upper limit for the integral

the upper limit is where f(x)=1 but you chose 1 but thats wrong so it becomes wrong where you equalize
F(x0)-F(1)=0.5
it should be
F(x0)-F(x1)+1-x1=0,5

[Szharz]

X= over 9000!!!! Gief account nao den!! Pliiiiiiiz

[ReidE96]

Ah, of course, that would be it. Thanks, blubb. It annoys me if I get something wrong and can't figure out where it went balls up. Not that I ever really intend on going back over it, I just like to know XD