Let $b, c \in \mathbb N$ with $2\le b <c$.
We define the function $f : [0, 1] \to [0, 1]$ such that: \begin{align} x = (0.\ a_1 a_2 a_3...)_b &\mapsto f(x) = (0.\ a_1 a_2 a_3...)_c \\ f(1) &\mapsto 1 \end{align} where $(0.\ a_1a_2a_3...)_b$ denotes a decimal number in base $b$ (similarly for $c$), that is: $$ (0.\ a_1a_2a_3...)_b = \sum_k \frac{a_k}{b^k} \qquad a_k \in\{0,1,\dots,b-1\} $$
Question: calculate $\int_0^1 f(x)\ dx $.
From $b < c$, I deduced that $f(x) < x,\ \forall x \in [0, 1)$, but that's all I have.
Divide $[0,1]$ into $b$ regions of width $b^{-1}$. Then, in the $k^\text{th}$ region, the first term in the $b$-expansion is $k-1$. Now, consider $$ \int_{(k-1)b^{-1}}^{kb^{-1}}f(x)dx. $$ Now, multiply this by $c$, you get \begin{align*} c\int_{(k-1)b^{-1}}^{kb^{-1}}f(x)dx&=\int_{(k-1)b^{-1}}^{kb^{-1}}((k-1)+f(b(x-(k-1)b^{-1})))dx\\ &=\frac{k-1}{b}+\int_{(k-1)b^{-1}}^{kb^{-1}}f(bx-k+1)dx \end{align*} Then, using the $u$-substitution $u=bx-k+1$, we get $du=bdx$, the lower limit of the integral is $0$ and the upper limit is $1$. Therefore, $$ c\int_{(k-1)b^{-1}}^{kb^{-1}}f(x)dx=\frac{k-1}{b}+\frac{1}{b}\int_0^1f(u)du. $$
From here, we can proceed as follows: $$ \int_0^1f(x)dx=\sum_{k=1}^{b}\int_{(k-1)b^{-1}}^{kb^{-1}}f(x)dx=\frac{1}{bc}\sum_{k=1}^b\left[(k-1)+\int_0^1f(x)dx\right]. $$ Let $\int_0^1f(x)dx=S$, then $$ S=\frac{1}{c}S+\frac{b-1}{2c}. $$ From this, we get that $$ S\left(\frac{c-1}{c}\right)=\frac{b-1}{2c} $$ or that $$ S=\frac{b-1}{2(c-1)}. $$ This answer passes the "sanity check" because when $b=c$, then $f(x)=x$ and the area of the triangle determined by $f(x)$ is $\frac{1}{2}$.
The idea behind the first part is that if $$ x=\sum_{k=1}^\infty \frac{a_k}{b^k}=0.a_1a_2a_3a_4\cdots_b $$ then $$ f(x)=\sum_{k=1}^\infty\frac{a_k}{c^k}=0.a_1a_2a_3a_4\cdots_c $$ Then $$ cf(x)=a_1+\sum_{k=1}^\infty\frac{a_{k+1}}{c^k}=a_1.a_2a_3a_4\cdots_c $$ Observe also that $$ bx=a_1+\sum_{k=1}^\infty\frac{a_{k+1}}{b^k}=a_1.a_2a_3a_4\cdots_b $$ Combining all of this, we see that $$ f(bx-a_1)=0.a_2a_3a_4\cdots_c. $$ Therefore, $$ cf(x)=a_1+f(bx-a_1). $$ Since $a_1$ varies between $0$ and $b-1$, every point in the $k$-th interval $((k-1)b^{-1},kb^{-1})$ has $a_1=k-1$.