Integral and mean value theorem question

If $\lambda(x) = -1$ for some $x \in X$, then no. Otherwise,

Let $$r = \frac 1{1+\lambda}$$ Your integral equation becomes $$\begin{align}\int_X r(x)f(x)+(1-r(x))g(x)\,dx &= r(x_0)\int_Xf(x)\,dx + (1-r(x_0))\int_xg(x)\,dx\\\int_Xg(x)\,dx + \int_Xr(x)(f(x) - g(x))\,dx &=\int_Xg(x)\,dx + r(x_0)\int_Xf(x) - g(x)\,dx \\\int_Xr(x)(f(x) - g(x))\,dx &= r(x_0)\int_Xf(x) - g(x)\,dx \end{align}$$

Since $r$ is continuous on the compact set $X$, it has a minimum value $m$ and maximum value $M$. So $$m\int_Xf(x) - g(x)\,dx\le\int_Xr(x)(f(x) - g(x))\,dx\le M\int_Xf(x) - g(x)\,dx$$ $$m \le \dfrac{\int_Xr(x)(f(x) - g(x))\,dx}{\int_Xf(x) - g(x)\,dx} \le M$$

Since $X$ is path-connected, there is some curve $\gamma$ with $r(\gamma(0)) = m$ and $r(\gamma(1)) = M$. By the intermediate value theorem, there is some $t \in [0,1]$ with $r(\gamma(t)) = \frac{\int_Xr(x)(f(x) - g(x))\,dx}{\int_Xf(x) - g(x)\,dx}$.

Setting $x_0 = \gamma(t)$ gives the result you are after.