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

The integral from 1 to 4 of y^3+1/(4y^3) is 8175/128

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

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

\frac{8175}{128}

Decimal

63.8671875

Solution steps

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

\int_{1}^{4} y^{3} d y = \frac{255}{4}

\int_{1}^{4} \frac{1}{4 y ^{3}} d y = \frac{15}{128}

= \frac{255}{4} + \frac{15}{128}

Simplify

\frac{255}{4} + \frac{15}{128} : \frac{8175}{128}

= \frac{8175}{128}

\frac{\partial}{\partial x} (\frac{1}{x x})

\frac{d}{d x} (e^{x - 5} - x)

f (x) = \frac{( x ^{2} )}{( 4 + x )}

\int_{1}^{4} t^{2} ln (4 t) d t

y^{^{'}} + 6 y = t + e^{- 4 t}

What is the integral from 1 to 4 of y^3+1/(4y^3) ?
The integral from 1 to 4 of y^3+1/(4y^3) is 8175/128