How do you evaluate the definite integral $\int (2 x^{- 2} - 3) d x$ $int (2x^-2-3) dx$ from $[1, 2]$ $[1,2]$ ?

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Answer 1 · 2017-03-03T01:49:56

$- 2$ $-2$

$\int_{1}^{2} (2 x^{- 2} - 3) d x = \int_{1}^{2} (2 x^{- 2}) d x - \int_{1}^{2} 3 d x$ $int_1^2 (2x^-2-3)dx = int_1^2 (2x^-2)dx - int_1^2 3dx$

Applying the power rule to each integral:

$= \frac{2 x^{- 1}}{- 1}]_{1}^{2} - 3 x]_{1}^{2}$ $= (2x^-1)/-1]_1^2 - 3x]_1^2$

$= \frac{- 2}{x}]_{1}^{2} - 3 x]_{1}^{2}$ $= (-2)/x]_1^2 - 3x]_1^2$

$= - 1 - (- 2) - (6 - 3)$ $=-1-(-2)-(6-3)$

$= 1 - 3 = - 2$ $=1-3 =-2$

How do you evaluate the definite integral ∫(2x−2−3)dxint (2x^-2-3) dx from [1,2][1,2]?