integrate x^2 between 0 and 1

Let's compute the integral:

\int_{0}^{1} x^{2} d x

ⓘ For constant

n

with

n \neq -1

\frac{x^{n + 1}}{n + 1}

is an antiderivative of

x^{n}

\int x^{2} d x

\frac{x^{3}}{3}

▷

\frac{x^{3}}{3}

Let's evaluate for the interval

[0, 1]

\frac{1^{3}}{3} - \frac{0^{3}}{3}

\frac{1}{3} + 0

ⓘ Adding 0 changes nothing:

\frac{1}{3} + 0

\frac{1}{3}

▶Integral:

\frac{1}{3}

The numeric value of:

\frac{1}{3}

equals:

▷

\approx 0.3333

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