Solution assignment 08 Integration by substitution

Solve:

$\displaystyle\int_{0}^{1}[(3x+1)^2-(3x-1)^2]dx$

We can integrate this integral term by term, but there is an easier way. We apply some special functions:

$(3x+1)^2-(3x-1)^2=12x$

Solving the integral is quite easy now:

$\displaystyle\int_{0}^{1}[(3x+1)^2-(3x-1)^2]dx=\displaystyle\int_{0}^{1}{12x}dx=6$