Question

Let f is a differentiable function such that f(x)=x^2 + integral of e^-tf(x-t)dt where t goes from 0 to x then
f(3) =?

Answer 1

@hartnn

Answer 2

Actually what i did was first applied Leibnitz's rule and diff f(x).
Solving that i got this
F'(x)=2x+e^-xf(0)
Now f(0)=0 from the equation of f(x) given in the question.
So if we just integrate then i got
F(x)=x^2
So ans should be 9.
Bt it's not ryt..
Where am i wrong??
:(

Answer 3

have you tried using the property :
\(\large \int_a^b f(z)dz = \int_a^b f(a+b-z)dz\)

Answer 4

I didn't..
I just diff the expression n again integrated it!

Answer 5

Can u tell is this right .
F(0)=0

Answer 6

thats right \(\int_0^0f(x)\,dx = 0\)

Answer 7

So ans is not coming if i do it here..
Like first of all if i diff the expression then i m getting f'(x)=2x+e^-x f(0)
That means f'(x)=2x
Now just integrate it..
We will get f(x)=x^2 +c
Again c=0 for f(0)=0

Answer 8

That means f(3)=9.

Answer 9

That means f(3)=9.

Answer 10

your differentiation is wrong

Answer 11

fundemental theorem of arithmetic 2 gives :
\[\dfrac{d}{d\color{red}{x}} \int\limits_a^{\color{red}{x}} g(t)\,dt = g(\color{red}{x})\]

Answer 12

you have applied it wrong - your integrand contains `x` terms too

Answer 13

Okay ! So we can apply this Leibnitz rule only for
Integral of g(t)dt containing no other variable..

Answer 14

\[f(\color{red}{x}) = \color{red}{x}^2+\int\limits_0^{\color{red}{x}} e^{-t}f(\color{red}{x}-t)\,dt\]

Answer 15

yes, first try and pull the `x` terms out the integrand

Answer 16

That can be only done as suggested by hartnn's method!!

Answer 17

yes, do it

Answer 18

Okay!

Answer 19

or you can also make a substitution u =x-t

Answer 20

I got d ans which is 18..!

Answer 21

Thanks !! Both of you

Answer 22

Awesome!