Let f(x)=x^3+x+1 for real x and let g=f^-1 be its inverse function. The value of ∫1^3 g(y) dy is

$f'(x)=3x^2+1>0$, so $f$ is strictly increasing and $g$ exists. Since $f(0)=1$ and $f(1)=3$, the region under $y\mapsto g(y)$ for $1\le y\le3$ is reflected to the region between $x=0,x=1$ and above the graph of $f$ in the rectangle. Since $g$ is the reflection of $f$ in $y=x$, the two areas together fill the rectangle $[0,1]\times[1,3]$ plus the part under $y=1$: precisely, $\int_1^3 g(y)dy+\int_0^1 f(x)dx=1\cdot3-0\cdot1=3$ (integration by parts for inverse functions). Now $\int_0^1(x^3+x+1)dx=\frac14+\frac12+1=\frac74$. Hence $\int_1^3 g=3-\frac74=\frac54$. The option $\frac74$ is $\int_0^1f$ itself, and $\frac{13}{4}$ results from adding instead of subtracting.