What is the integral from 1 to 2 of (y^4)/8+1/(4y^2) ?

The integral from 1 to 2 of (y^4)/8+1/(4y^2) is 9/10

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integral from 1 to 2 of (y^4)/8+1/(4y^2)

\int_{1}^{2} \frac{y ^{4}}{8} + \frac{1}{4 y ^{2}} d y

\frac{9}{10}

Decimal

0.9

Solution steps

\int_{1}^{2} \frac{y ^{4}}{8} + \frac{1}{4 y ^{2}} 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} \frac{y ^{4}}{8} d y + \int_{1}^{2} \frac{1}{4 y ^{2}} d y

\int_{1}^{2} \frac{y ^{4}}{8} d y = \frac{31}{40}

\int_{1}^{2} \frac{1}{4 y ^{2}} d y = \frac{1}{8}

= \frac{31}{40} + \frac{1}{8}

Simplify

\frac{31}{40} + \frac{1}{8} : \frac{9}{10}

= \frac{9}{10}

What is the integral from 1 to 2 of (y^4)/8+1/(4y^2) ?
The integral from 1 to 2 of (y^4)/8+1/(4y^2) is 9/10