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

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

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integral from 0 to 1 of sqrt(t^4+t^2)

\int_{0}^{1} t^{4} + t^{2} d t

\frac{2 2 - 1}{3}

Decimal Notation

0.60947 \dots

Solution steps

\int_{0}^{1} t^{4} + t^{2} d t

Simplify

t^{4} + t^{2} : t^{2} t^{2} + 1

= \int_{0}^{1} t^{2} t^{2} + 1 d t

t^{2} = (t),

assuming

t \geq 0

= \int_{0}^{1} t t^{2} + 1 d t

Apply u-substitution

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

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

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{1}^{2}

= [\frac{u ^{3}}{3}]_{1}^{2}

Compute the boundaries:

\frac{2 2}{3} - \frac{1}{3}

= \frac{2 2}{3} - \frac{1}{3}

Simplify

= \frac{2 2 - 1}{3}

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