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

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

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

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

\frac{16}{63}

Decimal Notation

0.25396 \dots

Solution steps

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

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

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

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

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

Simplify

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

= \frac{16}{63}

\frac{d y}{d x} + 4 x y = x e^{x^{2}}

x^{2} \cdot e^{x}

x \to - 5 lim (\frac{3 x + 8}{x + 5})

4 t

x \to 14 - lim (ln (14 - x))

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