$%27)I=\displaystyle\int\limits^1_0 x^2f"(x) \, dx\\ %u=x^2 \Rightarrow du=2xdx\\ %dv= f"(x) \, dx \Rightarrow v=f'(x)\\ %\displaystyle\int\limits^1_0 x^2f"(x) \, dx\\ %=x^2f'(x)\Bigg\vert^1_0-\displaystyle\int\limits^1_0 f'(x)2x \, dx\\ %=f'(1)-\displaystyle\int\limits^1_0 f'(x)2x \, dx\\ %=2e^2-I_2\\ %I_2=\displaystyle\int\limits^1_0 f'(x)2x \, dx\\ %u=2x \Rightarrow du=2dx\\ %dv= f'(x) \, dx \Rightarrow v=f(x)\\ %I_2=\displaystyle\int\limits^1_0 f'(x)2x \, dx\\ %=2xf(x)\Bigg\vert^1_0-\displaystyle\int\limits^1_0 f(x)dx \, dx%$
$28)f(x)-xf'(x)=-2x^3\\ \Leftrightarrow \dfrac{xf'(x)-f(x)}{x^2}=2x\\ \Leftrightarrow \displaystyle\int \dfrac{xf'(x)-f(x)}{x^2} \, dx=\displaystyle\int 2x \, dx\\ \Leftrightarrow \dfrac{f(x)}{x}=x^2+C\\ f(1)=1 \Rightarrow C=0\\ \dfrac{f(x)}{x}=x^2\\ \Rightarrow f(x)=x^3\\ \displaystyle\int\limits^1_0 f(x) \, dx\\ =\displaystyle\int\limits^1_0 x^3 \, dx\\ =\left(\dfrac{x^4}{4}\right)\Bigg\vert^1_0\\ =\dfrac{1}{4}$
