## Calculus: Early Transcendentals 8th Edition

Original Equation: $\int(t(1-t)^{2})dt$ on the interval $[1,-1]$ First, we have to expand the equation before we evaluate the integral. Using basic math and the FOIL method, we seethe equationis simplified as shown: $\int(t(1-t)^{2})dt=$ $\int(t(t^{2}-2t+1)dt=$ $\int(t)^{3}-2t^{2}+t)dt$ To solve this integral, we first need to find the anti-derivative. The anti-derivative of $x^{n}$ is found through the equation $\frac{x^{n+1}}{n+1}$. by applying this formula to each term in the equation, we see that the final anti-derivative is $\frac{t^{4}}{4}-\frac{2t^{3}}{3}+\frac{t^{2}}{2}$ Now that we have this equation, we simply subtract the bottom range from the upper range. Our range is $[1,-1]$, so we plug 1 and -1 into the anti derivative and the difference of the two is our final answer. $(\frac{(1)^{4}}{4}-\frac{2(1)^{3}}{3}+\frac{(1)^{2}}{2})$ +$(\frac{(1)^{4}}{4}-\frac{2(-1)^{3}}{3}+\frac{(-1)^{2}}{2})$$= \frac{-4}{3}$