# Solving the equation of the type $g\left( x \right) = \int\limits_0^x {f\left( t \right)dt}$

Let $$f:\left[ {0,1} \right] \to R$$ be a differentiable function. Let $$g\left( x \right) = \int\limits_0^x {f\left( t \right)dt}$$ with $$g\left( 1 \right) = 0$$ . Which of these equations must have at least one solution for x in the interval $$(0,1)$$ ?

(A). $$g\left( x \right) = f\left( x \right)$$

(B). $$xg\left( x \right) = \left( {1 - x} \right)f\left( x \right)$$

(C). $$f\left( x \right) = f'\left( x \right)g\left( x \right)$$

(D). $$f\left( x \right) = xg\left( x \right)$$

This question has one or more than one correct

My approach is as follow

$$g\left( 0 \right) = \int\limits_0^0 {f\left( t \right)dt} = 0$$

Given $$g(1)=0$$

On differentiating we get $$g'\left( x \right) = f\left( x \right)$$, hence $$g'\left( c \right) = f\left( c \right)$$ where $$c\in (0,1)$$. How we will use the Rolle's theorem for checking and verifying each option choice.

• $f=g=0$ is a solution for all this equations, perhaps the question here is about non trivial solution ? Aug 24 at 9:37
• I think all the options are correct. By the way, @Hamza, I don't think the question asks about existence of such functions. I think the question asks about the solutions of the given equations for any differentiable function $f$ and $g$ as defined. Aug 24 at 10:22

I think all the options have at least one solution in $$(0,1)$$.

First of all, $$g'(x)=f(x)$$.

For option A,

Let $$F(x)=e^{-x}g(x)$$.

$$\implies F(0)=F(1)=0$$

$$\implies \exists c \in (0,1)$$ for which $$F'(x)=0$$

And $$F'(x)=e^{-x}g'(x) -e^{-x}g(x)=e^{-x}(f(x)-g(x))=0$$

$$\implies f(x)-g(x)=0$$ has a solution in $$(0,1)$$.

For option B,

Let $$G(x)=e^{x+\log(1-x)}g(x)$$

$$G(0)=G(1)=0\implies G'(c)=0$$ for some $$c\in (0,1)$$.

And $$G'(x)=e^{x+\log(1-x)}g'(x)+e^{x+\log(1-x)}\left(1-\frac{1}{1-x}\right)g(x)=0$$

$$\implies e^{x+\log(1-x)}\left(f(x)-\frac{xg(x)}{1-x}\right)=0$$

$$\implies (1-x)f(x)=xg(x)$$ has a solution in $$(0,1)$$.

For option C,

Let $$H(x)=e^{-f(x)}g(x)$$

So $$H(0)=H(1)=0 \implies H'(c)=0$$ for some $$c\in (0,1)$$

And $$H'(x)=e^{-f(x)}f(x)-e^{-f(x)}f'(x)g(x)=0$$

$$\implies e^{-f(x)}(f(x)-f'(x)g(x))=0$$

$$\implies f(x)=f'(x)g(x)$$ has a solution in $$(0,1)$$.

For option D,

Let $$I(x)=e^{-\frac{x^2}{2}}g(x)$$

$$I(0)=I(1)=0 \implies I'(c)=0$$ for some $$c\in(0,1)$$.

And $$I'(x)=e^{-\frac{x^2}{2}}f(x)-xe^{-\frac{x^2}{2}}g(x)=0$$

$$\implies e^{-\frac{x^2}{2}}(f(x)-xg(x))=0$$

$$\implies f(x)=xg(x)$$ has a solution in $$(0,1)$$.