evaluate $int_{-2}^{3}(4 - 2x^{3})dx$

Explanation:

Step1: Expand the integrand

We expand $(4 - 2x^{3})$ to get $4-2x^{3}$. Then $\int_{-2}^{3}(4 - 2x^{3})dx=\int_{-2}^{3}4dx-\int_{-2}^{3}2x^{3}dx$.

Step2: Integrate term - by - term

For $\int_{-2}^{3}4dx$, using the power - rule $\int kdx=kx + C$ ($k$ is a constant), we have $4x\big|{-2}^{3}=4\times(3-( - 2))=4\times5 = 20$. For $\int{-2}^{3}2x^{3}dx$, using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $2\times\frac{x^{4}}{4}\big|{-2}^{3}=\frac{1}{2}x^{4}\big|{-2}^{3}=\frac{1}{2}(3^{4}-(-2)^{4})=\frac{1}{2}(81 - 16)=\frac{1}{2}\times65 = 32.5$.

Step3: Calculate the result

$\int_{-2}^{3}(4 - 2x^{3})dx=20-32.5=-12.5$.

Answer:

$-12.5$