91Ó°ÊÓ

First, let's write down the original integral: $$ \int_{-2}^{2} \sqrt[5]{x^2} \,dx. $$

Now, we make the substitution \(t = x^{\frac{2}{5}}\), so \(x = t^{\frac{5}{2}}\). Let's evaluate the derivatives of \(x\) with respect to \(t\): $$\frac{dx}{dt} = \frac{5}{2} t^{\frac{3}{2}}.$$ Now, we need to change the integral limits according to the new variable, \(t\): $$t_{-2} = (-2)^{\frac{2}{5}} = 2^{\frac{2}{5}}, \quad t_{2} = (2)^{\frac{2}{5}} = 2^{\frac{2}{5}}.$$

Now we can change the integral in terms of \(t\), which gives us: $$ \int_{2^{\frac{2}{5}}}^{2^{\frac{2}{5}}} \sqrt[5]{(t^{\frac{5}{2}})^2} \, \frac{5}{2} t^{\frac{3}{2}} \, dt. $$

Now, let's simplify the integral: $$ \frac{5}{2}\int_{2^{\frac{2}{5}}}^{2^{\frac{2}{5}}} t^{\frac{3}{2}} \, t^{\frac{5}{2}} \, dt. $$ Combining the powers of \(t\), we get: $$ \frac{5}{2}\int_{2^{\frac{2}{5}}}^{2^{\frac{2}{5}}} t^4 \, dt. $$

At this point, we can observe that the limits of integration are the same, namely \(2^{\frac{2}{5}}\). This is the mistake in our substitution, because we lose the information about the negative side of the integral when we substitute \(t = x^{\frac{2}{5}}\) since both \((-2)^{\frac{2}{5}}\) and \((2)^{\frac{2}{5}}\) map to the same value. When we substituted \(t = x^{\frac{2}{5}}\), we lost information about the negative values of \(x\). The fifth root of a number squared will always be positive, so the substitution does not account for the negative values of the original integral, creating an incorrect result. To avoid this error while solving this integral, we should either split the original integral into two parts (positive side and negative side), or find an appropriate substitution that doesn't lose information about the negative side.