Ta có: \(x^4-\left(-x+2\right)^2=\left(x^2-x+2\right)\left(x^2+x-2\right)\)
\(=\left(x^2-x+2\right)\left(x+2\right)\left(x-1\right)\le0\) ; \(\forall x\in\left(0;1\right)\)
\(\Rightarrow\int\limits^1_0\left|x^4-\left(-x+2\right)^2\right|dx=\int\limits^1_0\left[\left(-x+2\right)^2-x^4\right]dx\)
\(=\int\limits^1_0\left(x^2-4x+4-x^4\right)dx=\dfrac{32}{15}\)