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 ∫−31(3t2 + 7t)dt. enter your answer as an exact fraction if necessary. provide your answer below:

current objective evaluate definite integrals with the fundamental theorem of calculus for functions with positive integer exponents question evaluate ∫−31(3t2 + 7t)dt. enter your answer as an exact fraction if necessary. provide your answer below:

Answer

Explanation:

Step1: Find antiderivative

The antiderivative of $3t^{2}$ is $3\times\frac{t^{3}}{3}=t^{3}$ (using power - rule $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C,n\neq - 1$), and the antiderivative of $7t$ is $7\times\frac{t^{2}}{2}=\frac{7t^{2}}{2}$. So the antiderivative of $3t^{2}+7t$ is $F(t)=t^{3}+\frac{7t^{2}}{2}$.

Step2: Apply the Fundamental Theorem of Calculus

$\int_{-3}^{1}(3t^{2}+7t)dt=F(1)-F(-3)$. $F(1)=1^{3}+\frac{7\times1^{2}}{2}=1+\frac{7}{2}=\frac{2 + 7}{2}=\frac{9}{2}$. $F(-3)=(-3)^{3}+\frac{7\times(-3)^{2}}{2}=-27+\frac{7\times9}{2}=-27+\frac{63}{2}=\frac{-54 + 63}{2}=\frac{9}{2}$. Then $F(1)-F(-3)=\frac{9}{2}-\frac{9}{2}=0$.

Answer:

$0$