How do you evaluate the definite integral $\int (x^{2} - 9) d x$ $int (x^2-9)dx$ from [3,4]?

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Answer 1 · 2016-11-13T13:44:30

The answer is $= \frac{10}{3}$ $=10/3$

To evaluate the integral, we use

$\int x^{n} = \frac{x^{n + 1}}{n + 1} + C$ $intx^n=x^(n+1)/(n+1)+C$

$\int_{3}^{4} (x^{2} - 9) d x = \int_{3}^{4} x^{2} d x - \int_{3}^{4} 9 d x$ $int_3^4(x^2-9)dx= int_3^4x^2dx-int_3^4 9dx$

$= {[\frac{x^{3}}{3} - 9 x]}_{3}^{4}$ $=[x^3/3-9x ]_3^4$

$= (\frac{64}{3} - 36) - (9 - 27)$ $=(64/3-36)-(9-27)$

$= \frac{64}{3} - 36 - 9 + 27$ $=64/3-36-9+27$

$= \frac{10}{3}$ $=10/3$

How do you evaluate the definite integral ∫(x2−9)dxint (x^2-9)dx from [3,4]?