integral from 0 to 1 of \sqrt[4]{x} — step-by-step solution

Let's integrate:

\int_{0}^{1} \sqrt[4]{x} d x

We integrate

\sqrt[4]{x}

by applying:

\int x^{n} d x = \frac{x^{n + 1}}{n + 1}

ⓘ An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

\int \sqrt[4]{x} d x

\frac{\sqrt[4]{x^{5}}}{\frac{5}{4}}

\frac{\sqrt[4]{x^{5}}}{\frac{5}{4}}

ⓘ Dividing is multiplying by the reciprocal:

\frac{\sqrt[4]{x^{5}}}{\frac{5}{4}}

\sqrt[4]{x^{5}} \cdot \frac{4}{5}

\sqrt[4]{x^{5}} \cdot \frac{4}{5}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\sqrt[4]{x^{5}} \cdot \frac{4}{5}

\frac{\sqrt[4]{x^{5}} \cdot 4}{5}

▷

\frac{\sqrt[4]{x^{5}} \cdot 4}{5}

Let's evaluate for the interval

[0, 1]

\frac{4 \cdot \sqrt[4]{1^{5}}}{5} - \frac{4 \cdot \sqrt[4]{0^{5}}}{5}

ⓘ A product with a zero factor is zero:

4 \cdot 0

0

\frac{4}{5} + 0

ⓘ Adding 0 changes nothing:

\frac{4}{5} + 0

\frac{4}{5}

▶Integral:

\frac{4}{5}

The numeric value of:

\frac{4}{5}

is:

▷

0.8