A function that integral equals to the function itself, but not an exponential function

Question

I want to find some examples of function (continuous, differentiable, and easy) that $$ \int_0^1 f(x)\;dx = f(1), $$ where $f(x)>0$ for all $x\in[0,1]$ and $f(0)=1$.

Notes

It is obvious that there are infinitely many functions that satisfy the above identity because the function is restricted only for $x=0$ and $x=1$, not $x\in(0,1)$. For example, $f(x)=e^x-2x$ satisfies the above condition. However, I want the function to have an exponential form like $f(x)=e^{g(x)}$.

Answer 1

There is no solution of the form $e^{ax}$ because $$\int_0^1 e^{ax}dx=\frac{e^a-1}a=f(1)=e^a$$ has no other solution than $a=0$.

And unless $g$ is a logarithm,

$$e^{g(x)}$$ usually has no analytical antiderivative.

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Better luck with a polynomial:

$$\int_0^1\left(ax-1\right)^2\,dx=\frac{(a-1)^3+1}{3a}=f(1)=(a-1)^2$$

has the solution

$$a=\frac32.$$

Answer 2

Let $f(x)=Ax^2+Bx+1 \implies f(0)=1.$ $$\int_{0}^{1} f(x) dx=Ax^3/3+Bx^2/2+x |_{x=0}^{1}=f(1) \implies A/3+B/2+1=A+B+1$$ $$\implies -2A/3=B/2 \implies A/B=-3/4$$ Let $A=3/4, B=-1$, then $f(x)=3x^2/4-x+1>0, x\in R$