Solve Integration area I am confused why the answer for $$\int_0^{\infty}\left(x-\frac{1}{\lambda}\right)^2\lambda e^{-\lambda x}\ dx$$
  is $$\frac{1}{\lambda^2}$$
I get mine as $$\frac{5}{\lambda^2}$$
Official answer is as follows 

but I do not get the last part when it is $$\frac{-2}{\lambda^2}$$
 instead of $$\frac{2}{\lambda^2}$$
My working for the step before is as follows
f(x) = 2x-2 \quad
f'(x) = 2 \quad
g(x) = -(e^-λx)/λ \quad
g'(x)= e^-λx
Essentially, it means Area f(x)g(x) - Integral f'(x)g(x)dx. Why is it -2/λ when f'(x)g(x) = -2(e^-λx)/λ and I take -2/λ out of the integral to make it 
2/λ∫(e^-λx)20dx

The last integral is
$$-\frac{2}{\lambda} \int_0^{\infty} e^{-\lambda x}\; dx  = \left.\frac{-2}{\lambda} \frac{e^{-\lambda x}}{-\lambda} \right|_{0}^{\infty} = \frac{2}{\lambda^2} \left(e^{-\infty} - e^0 \right) = \frac{2}{\lambda^2}(0-1) = -\frac{2}{\lambda^2} .$$
