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 switch question what is the value of the definite integral ∫−33(3x3−2x2+x+1)dx? 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 switch question what is the value of the definite integral ∫−33(3x3−2x2+x+1)dx? enter your answer as an exact fraction if necessary. provide your answer below:

Answer

Explanation:

Step1: Find antiderivative

The antiderivative of $3x^{3}$ is $\frac{3}{4}x^{4}$, of $- 2x^{2}$ is $-\frac{2}{3}x^{3}$, of $x$ is $\frac{1}{2}x^{2}$ and of $1$ is $x$. So the antiderivative of $3x^{3}-2x^{2}+x + 1$ is $F(x)=\frac{3}{4}x^{4}-\frac{2}{3}x^{3}+\frac{1}{2}x^{2}+x$.

Step2: Apply fundamental theorem

By the fundamental theorem of calculus $\int_{a}^{b}f(x)dx=F(b)-F(a)$. Here $a=-3$ and $b = 3$. $F(3)=\frac{3}{4}(3)^{4}-\frac{2}{3}(3)^{3}+\frac{1}{2}(3)^{2}+3=\frac{3}{4}\times81-\frac{2}{3}\times27+\frac{1}{2}\times9 + 3=\frac{243}{4}-18+\frac{9}{2}+3$. $F(-3)=\frac{3}{4}(-3)^{4}-\frac{2}{3}(-3)^{3}+\frac{1}{2}(-3)^{2}+(-3)=\frac{3}{4}\times81+\frac{2}{3}\times27+\frac{1}{2}\times9-3=\frac{243}{4}+18+\frac{9}{2}-3$. $F(3)-F(-3)=\left(\frac{243}{4}-18+\frac{9}{2}+3\right)-\left(\frac{243}{4}+18+\frac{9}{2}-3\right)$. The $\frac{243}{4}$ and $\frac{9}{2}$ terms cancel out. $F(3)-F(-3)=(-18 + 3)-(18-3)=-15 - 15=-30$.

Answer:

$-30$