What is the integral from-1 to-2 of (16-x^3)/(x^3) ?

The integral from-1 to-2 of (16-x^3)/(x^3) is 7

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integral from-1 to-2 of (16-x^3)/(x^3)

\int_{- 1}^{- 2} \frac{16 - x ^{3}}{x ^{3}} d x

7

Solution steps

\int_{- 1}^{- 2} \frac{16 - x ^{3}}{x ^{3}} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{- 1} \frac{16 - x ^{3}}{x ^{3}} d x

Expand

\frac{16 - x ^{3}}{x ^{3}} : \frac{16}{x ^{3}} - 1

= - \int_{- 2}^{- 1} \frac{16}{x ^{3}} - 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int_{- 2}^{- 1} \frac{16}{x ^{3}} d x - \int_{- 2}^{- 1} 1 d x)

\int_{- 2}^{- 1} \frac{16}{x ^{3}} d x = - 6

\int_{- 2}^{- 1} 1 d x = 1

= - (- 6 - 1)

Simplify

= 7

\int 5 sin^{3} (5 x) cos^{3} (5 x) d x

\frac{\partial}{\partial x} (x^{2} y + e^{x y})

\frac{\partial ^{3}}{\partial x \partial y \partial z} (e^{x y z^{9}})

\frac{d}{d t} (sin (t))

3 x y^{^{'}} - 15 y = x^{- 4}, y (1) = - 2

What is the integral from-1 to-2 of (16-x^3)/(x^3) ?
The integral from-1 to-2 of (16-x^3)/(x^3) is 7