Question

integral definida de 1 a 4 √t (1+t) dt

Answer 1

$\int\limits^4_1 {\sqrt{t}(1+t)}} \, dt = \\\\\int\limits^4_1 {t^{\frac12}(1+t)}} \, dt =\\\\ \int\limits^4_1 {t^\frac12+t^{1+\frac12}}} \, dt =\\\\\int\limits^4_1 {t^\frac12+t^{\frac32}}} \, dt =\\\\\left[\dfrac{t^{1+\frac12}}{1+\frac12}+\dfrac{t^{1+\frac32}}{1+\frac32}\right]^4_1=\\\\\left[\dfrac{t^\frac32}{\frac32}+\dfrac{t^\frac52}{\frac52}\right]^4_1=\\\\\left[\dfrac{4^\frac32}{\frac32}+\dfrac{4^\frac52}{\frac52}\right]-\left[\dfrac{1^\frac32}{\frac32}+\dfrac{1^\frac52}{\frac52}\right]=\\\\\dfrac{16}{3}+\dfrac{64}{5}-\dfrac{2}{3}-\dfrac{2}{5}=\\\\\dfrac{14}{3}+\dfrac{62}{5}=\dfrac{256}{15}$