Calculus: Early Transcendentals 8th Edition

Original Equation: $\int$ $(x^{2}-3)dx$ on the interval $[2,1]$ 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 anti-derivative is $\frac{4x^{4}}{4}-\frac{3x^{3}}{3}+\frac{2x^{2}}{2}$. We can cancel out the numerators with the denominators to get $x^{4}-x^{3}+x^{2}$ Now that we have this equation, we simply subtract the bottom range from the upper range. Our range is $[2,1]$, so we plug 2 and 1 into the anti derivative and the difference of the two is our final answer. $(2^{4}-2^{3}+2^{2})$-$(1^{4}-1^{3}+1^{2})$ $= 11$