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

The integral from 1 to 3 of (y^4+1/(4y^4)) is 39269/810

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

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

\frac{39269}{810}

Decimal

48.48024 \dots

Solution steps

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

\int_{1}^{3} y^{4} d y = \frac{242}{5}

\int_{1}^{3} \frac{1}{4 y ^{4}} d y = \frac{13}{162}

= \frac{242}{5} + \frac{13}{162}

Simplify

\frac{242}{5} + \frac{13}{162} : \frac{39269}{810}

= \frac{39269}{810}

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = a, y^{^{'}} (0) = b

\int_{3}^{7} 1 d x

\int_{0}^{6} \frac{1}{18} x d x

\frac{d}{d x} (x π)

\frac{1}{( s + 3 ) ^{2}}

What is the integral from 1 to 3 of (y^4+1/(4y^4)) ?
The integral from 1 to 3 of (y^4+1/(4y^4)) is 39269/810