To evaluate the integral 0 (t1/3 − t2/3) dt −1 , first use the property of definite integrals, that is, b (f(x) + g(x)) dx a = b f(x) dx a + b g(x) dx. a Therefore, 0 (t1/3 − t2/3) dt −1 = 0 t1/ dt −1 ---Select--- 0 t /3 dt. −1