## Calculus: Early Transcendentals 8th Edition

$\displaystyle\int\limits_0^2(\dfrac{4}{5}t^{3}-\dfrac{3}{4}t^{2}+\dfrac{2}{5}t)dt=2$
$\displaystyle\int\limits_0^2(\dfrac{4}{5}t^{3}-\dfrac{3}{4}t^{2}+\dfrac{2}{5}t)dt$ Integrate each term separately: $\displaystyle\int\limits_0^2\dfrac{4}{5}t^{3}dt-\displaystyle\int\limits_0^2\dfrac{3}{4}t^{2}dt+\displaystyle\int\limits_0^2\dfrac{2}{5}tdt=...$ Take the constants out of the integral and continue integrating: $...=\displaystyle\dfrac{4}{5}\int\limits_0^2t^{3}dt-\displaystyle\dfrac{3}{4}\int\limits_0^2t^{2}dt+\displaystyle\dfrac{2}{5}\int\limits_0^2tdt=...$ $...=\dfrac{4}{5}(\dfrac{1}{4})t^{4}-\dfrac{3}{4}(\dfrac{1}{3})t^{3}+\dfrac{2}{5}(\dfrac{1}{2})t^{2}\Big|_0^2=...$ $...=\dfrac{1}{5}t^{4}-\dfrac{1}{4}t^{3}+\dfrac{1}{5}t^{2}\Big|_0^2=...$ Use the second part of the fundamental theorem of calculus to get the result: $...=\Big[\dfrac{1}{5}(2)^4-\dfrac{1}{4}(2)^{3}+\dfrac{1}{5}(2)^2\Big]-\Big[\dfrac{1}{5}(0)^4-\dfrac{1}{4}(0)^{3}+\dfrac{1}{5}(0)^2\Big]=...$ $...=\dfrac{1}{5}(16)-\dfrac{1}{4}(8)+\dfrac{1}{5}(4)=\dfrac{16}{5}-2+\dfrac{4}{5}=2$