current objective evaluate definite integrals with the fundamental theorem of calculus for functions with…

current objective evaluate definite integrals with the fundamental theorem of calculus for functions with positive integer exponents question evaluate the definite integral below. enter your answer as an exact fraction if necessary. ∫−3−1(t3 + 6t2 + 9t + 9)dt

current objective evaluate definite integrals with the fundamental theorem of calculus for functions with positive integer exponents question evaluate the definite integral below. enter your answer as an exact fraction if necessary. ∫−3−1(t3 + 6t2 + 9t + 9)dt

Answer

Explanation:

Step1: Find antiderivative

Use power - rule $\int t^n dt=\frac{t^{n + 1}}{n+1}+C$ ($n\neq - 1$). The antiderivative of $t^3+6t^2 + 9t+9$ is $\frac{t^{4}}{4}+6\times\frac{t^{3}}{3}+9\times\frac{t^{2}}{2}+9t=\frac{t^{4}}{4}+2t^{3}+\frac{9t^{2}}{2}+9t$.

Step2: Apply the Fundamental Theorem of Calculus

$\left[\frac{t^{4}}{4}+2t^{3}+\frac{9t^{2}}{2}+9t\right]_{-3}^{-1}$ $=\left(\frac{(-1)^{4}}{4}+2\times(-1)^{3}+\frac{9\times(-1)^{2}}{2}+9\times(-1)\right)-\left(\frac{(-3)^{4}}{4}+2\times(-3)^{3}+\frac{9\times(-3)^{2}}{2}+9\times(-3)\right)$ $=\left(\frac{1}{4}-2+\frac{9}{2}-9\right)-\left(\frac{81}{4}-54+\frac{81}{2}-27\right)$ $=\left(\frac{1 - 8+18 - 36}{4}\right)-\left(\frac{81-216 + 162-108}{4}\right)$ $=\frac{-25}{4}-\frac{-81}{4}$ $=\frac{-25 + 81}{4}=\frac{56}{4}=14$

Answer:

$14$