Question

Use Part II of the Fundamental Theorem of Calculus to evaluate the definite integral

Answer 1

`\begin{gathered} \int (4x^5-7x^2)/(x^3)dx=\int (4x^5)/(x^3)dx-\int (7x^2)/(x^3)dx \\ \int (4x^5-7x^2)/(x^3)dx=\int 4x^2dx-\int (7)/(x)dx \\ \int (4x^5-7x^2)/(x^3)dx=(4)/(3)x^3-7\ln (x) \end{gathered}`
`\begin{gathered} \int ^(-1)_(-2)(4x^5-7x^2)/(x^3)=((4)/(3)(-1)^3-7\ln (-1))-((4)/(3)(-2)^3-7\ln (-2)) \\ \int ^(-1)_(-2)(4x^5-7x^2)/(x^3)=(-(4)/(3)-7(i\cdot\pi))-((4)/(3)(-2)^3-7(\log \mleft(2\mright)+i\pi)) \\ \int ^(-1)_(-2)(4x^5-7x^2)/(x^3)=-(4)/(3)-7i\cdot\pi-(4)/(3)(-8)+7\log (2)+7i\cdot\pi \\ \int ^(-1)_(-2)(4x^5-7x^2)/(x^3)=-(4)/(3)+(32)/(3)+7\log (2) \\ \int ^(-1)_(-2)(4x^5-7x^2)/(x^3)=(28)/(3)+7\log (2) \\ \int ^(-1)_(-2)(4x^5-7x^2)/(x^3)=14.185 \end{gathered}`