What is the integral from 0 to 1 of sqrt(x^2-x^4) ?

The integral from 0 to 1 of sqrt(x^2-x^4) is 1/3

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integral from 0 to 1 of sqrt(x^2-x^4)

\int_{0}^{1} x^{2} - x^{4} d x

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

\int_{0}^{1} x^{2} - x^{4} d x

Simplify

x^{2} - x^{4} : x 1 - x^{2}

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

Apply u-substitution

: \int_{1}^{0} - u^{2} d u

= \int_{1}^{0} - u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - u^{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{0}^{1} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{1}

= - (- [\frac{u ^{3}}{3}]_{0}^{1})

Simplify

= [\frac{u ^{3}}{3}]_{0}^{1}

Compute the boundaries:

\frac{1}{3}

= \frac{1}{3}

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

What is the integral from 0 to 1 of sqrt(x^2-x^4) ?
The integral from 0 to 1 of sqrt(x^2-x^4) is 1/3