How did we get ( t3/3 )??

Dear Student,

A s \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C T h e r e f o r e, \int_{0}^{2} (t^{2} + 4) d t = \int_{0}^{2} (t^{2}) d t + \int_{0}^{2} (4) d t = {[\frac{t^{2 + 1}}{2 + 1}]}_{0}^{2} + {[\frac{4 t^{0 + 1}}{0 + 1}]}_{0}^{2} = {[\frac{t^{3}}{3}]}_{0}^{2} + {[4 t]}_{0}^{2} = [\frac{2^{3}}{3} - \frac{0^{3}}{3}] + [4 \times 2 - 4 \times 0] . . . . . . [a s {[f (x)]}_{a}^{b} = f (b) - f (a)] = \frac{8}{3} + 8 = \frac{32}{3}

Regards,