What is the integral from 0 to 1 of (y^6+y^8) ?

The integral from 0 to 1 of (y^6+y^8) is 16/63

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integral from 0 to 1 of (y^6+y^8)

\int_{0}^{1} (y^{6} + y^{8}) d y

\frac{16}{63}

Decimal Notation

0.25396 \dots

Solution steps

\int_{0}^{1} y^{6} + y^{8} 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^{6} d y + \int_{0}^{1} y^{8} d y

\int_{0}^{1} y^{6} d y = \frac{1}{7}

\int_{0}^{1} y^{8} d y = \frac{1}{9}

= \frac{1}{7} + \frac{1}{9}

Simplify

\frac{1}{7} + \frac{1}{9} : \frac{16}{63}

= \frac{16}{63}

x \to 1 + lim (\frac{1}{x} - 1)

y = 4 x, y = 6 x^{2}

\int 3 x^{2} a^{2} - x^{2} d x

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the integral from 0 to 1 of (y^6+y^8) ?
The integral from 0 to 1 of (y^6+y^8) is 16/63