What is the integral from 1 to 2 of (3-6x)/(x^3) ?

The integral from 1 to 2 of (3-6x)/(x^3) is -15/8

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integral from 1 to 2 of (3-6x)/(x^3)

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

- \frac{15}{8}

Decimal

- 1.875

Solution steps

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

Expand

\frac{3 - 6 x}{x ^{3}} : \frac{3}{x ^{3}} - \frac{6}{x ^{2}}

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

\int_{1}^{2} \frac{3}{x ^{3}} d x = \frac{9}{8}

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

= \frac{9}{8} - 3

Simplify

\frac{9}{8} - 3 : - \frac{15}{8}

= - \frac{15}{8}

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What is the integral from 1 to 2 of (3-6x)/(x^3) ?
The integral from 1 to 2 of (3-6x)/(x^3) is -15/8