$f(x)=\{\begin{array}{cl}8&\text{for }x\in [1,{3\over 2}]\\-{23\over 12}&\text{for }x\in ({3\over 2},2] \end{array}$ Then $\int_{1}^{2}f(x)dx={73\over 24}$ with $f(x)>x^{3}$ for $x\in [1,{3\over 2}]$ and $f(x)<x^{2}$ for $x\in ({3\over 2},2]$.

olorin wrote: $f(x)=\{\begin{array}{cl}8&\text{for }x\in [1,{3\over 2}]\\-{23\over 12}&\text{for }x\in ({3\over 2},2] \end{array}$ Then $\int_{1}^{2}f(x)dx={73\over 24}$ with $f(x)>x^{3}$ for $x\in [1,{3\over 2}]$ and $f(x)<x^{2}$ for $x\in ({3\over 2},2]$. how did u get the function and further it is given in the question to prove that there exists a $x_{0}$ which satisy the conditions and in ur case that diifers for the left hand side identity nand thje right hand side

pardesi wrote: how did u get the function and further it is given in the question to prove that there exists a $x_{0}$ which satisy the conditions and in ur case that diifers for the left hand side identity nand thje right hand side I simply searched $c,d\in\mathbb{R}$, $c$ large enough, $d$ small enough with $f(x)=\{\begin{array}{cl}c&\text{for }x\in [1,{3\over 2}]\\d&\text{for }x\in ({3\over 2},2] \end{array}$, such that $\int_{1}^{2}f(x)dx={c+d\over 2}={73\over 24}$. I therefore proved the problem incorrect.

Adding continuity as an hypothesis fixes it, I think.