## Calculus: Early Transcendentals 8th Edition

Original Equation: $\int$ $(2x-3)(4x^{2}+1)dx$ on the interval $[2,0]$ First, we have to expand the equation to make it easy to find the integral. We do this using the FOIL method. We see that after FOILing, the final equation becomes $\int$ $(8x^{3}-12x^{2}+2x-3)dx$ 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 $(8x^{4}/4-12x^{3}/3+2x^{2}/2-3x)$ which can be simplified to $(2x^{4}-4x^{3}+x^{2}-3x)$. Now that we have this equation, we simply subtract the bottom range from the upper range. Our range is $[2,0]$, so we plug 2 and 0 into the anti derivative and the difference of the two is our final answer. $(2(2)^{4}-4(2)^{3}+(2)^{2}-3(2))$ - $(2(0)^{4}-4(0)^{3}+(0)^{2}-3(0))$ $= -2$