What is the integral from 0 to 1 of (y^2+y^4) ?

The integral from 0 to 1 of (y^2+y^4) is 8/15

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integral from 0 to 1 of (y^2+y^4)

\int_{0}^{1} (y^{2} + y^{4}) d y

\frac{8}{15}

Decimal Notation

0.53333 \dots

Solution steps

\int_{0}^{1} y^{2} + 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_{0}^{1} y^{2} d y + \int_{0}^{1} y^{4} d y

\int_{0}^{1} y^{2} d y = \frac{1}{3}

\int_{0}^{1} y^{4} d y = \frac{1}{5}

= \frac{1}{3} + \frac{1}{5}

Simplify

\frac{1}{3} + \frac{1}{5} : \frac{8}{15}

= \frac{8}{15}

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

\int \frac{x}{y ^{2} + x ^{2}} d x

x \to 1 lim (\frac{x ^{3} - 1}{x - 6})

x^{e}

\int \frac{2 ^{x}}{x} d x

What is the integral from 0 to 1 of (y^2+y^4) ?
The integral from 0 to 1 of (y^2+y^4) is 8/15