Why not 101?

let $f(x)=(\log_{10}(x)-1)^2$ then $f'(x)=\dfrac{2(\log_{10}(x)-1)}{x\ln(x)}$ which is strictly decreasing for $x\leq10$ and strictly increasing for $x\geq10$ . Hence if we approximate the area under the curve between $f(x)$ as 2 triangles with bases from $x=1 \,\text{to}\, 10$ and $x=10 \,\text{to}\, 100$ , note $f(10) =0$ , we must get an over estimate. In doing this we find;

approximate area under $f(x)$ for $1\leq x\leq100$ = $\frac{1}{2}(9 \times 1 + 90 \times 1)$ = $49.5<50$ .