7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

$ = - \int_0^{1/2} {x\left( {x - \frac{1}{2}} \right) + \int_{1/2}^1 {x\left( {x - \frac{1}{2}} \right)} \,dx} $

$ = \int_0^{1/2} {\left( {\frac{1}{2}x - {x^2}} \right)\,dx + \int_{1/2}^1 {\left( {{x^2} - \frac{1}{2}x} \right)\,dx} } $

$ = \left( {\frac{{{x^2}}}{4} - \frac{{{x^3}}}{3}} \right)_0^{1/2} + \left( {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{4}} \right)_{1/2}^1$

$ = \left( {\frac{1}{{16}} - \frac{1}{{24}}} \right) + \left( {\frac{1}{3} - \frac{1}{4} + \frac{1}{{16}} - \frac{1}{{24}}} \right) = \frac{1}{8}.$