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

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

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integral from 1 to 2 of (8-3x)/(x^3)

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

\frac{3}{2}

Decimal

1.5

Solution steps

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

Expand

\frac{8 - 3 x}{x ^{3}} : \frac{8}{x ^{3}} - \frac{3}{x ^{2}}

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

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

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

= 3 - \frac{3}{2}

Simplify

3 - \frac{3}{2} : \frac{3}{2}

= \frac{3}{2}

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