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

The integral from 1 to 2 of (x^6+1)/(2x^3) is 33/16

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

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

\frac{33}{16}

Decimal Notation

2.0625

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

\frac{x ^{6} + 1}{x ^{3}} : x^{3} + \frac{1}{x ^{3}}

= \frac{1}{2} \cdot \int_{1}^{2} x^{3} + \frac{1}{x ^{3}} 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

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

\int_{1}^{2} x^{3} d x = \frac{15}{4}

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

= \frac{1}{2} (\frac{15}{4} + \frac{3}{8})

Simplify

\frac{1}{2} (\frac{15}{4} + \frac{3}{8}) : \frac{33}{16}

= \frac{33}{16}

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What is the integral from 1 to 2 of (x^6+1)/(2x^3) ?
The integral from 1 to 2 of (x^6+1)/(2x^3) is 33/16