What is the integral from 1 to 2 of (y^3+1/(4y^3)) ?

The integral from 1 to 2 of (y^3+1/(4y^3)) is 123/32

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integral from 1 to 2 of (y^3+1/(4y^3))

\int_{1}^{2} (y^{3} + \frac{1}{4 y ^{3}}) d y

\frac{123}{32}

Decimal

3.84375

Solution steps

\int_{1}^{2} y^{3} + \frac{1}{4 y ^{3}} d y

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} y^{3} d y + \int_{1}^{2} \frac{1}{4 y ^{3}} d y

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

\int_{1}^{2} \frac{1}{4 y ^{3}} d y = \frac{3}{32}

= \frac{15}{4} + \frac{3}{32}

Simplify

\frac{15}{4} + \frac{3}{32} : \frac{123}{32}

= \frac{123}{32}

What is the integral from 1 to 2 of (y^3+1/(4y^3)) ?
The integral from 1 to 2 of (y^3+1/(4y^3)) is 123/32