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

Use Part II Of The Fundamental Theorem Of Calculus To Evaluate The Definite Integral class=

Sagot :

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

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