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

The integral from 0 to 2pi of sqrt(θ^4+4θ^2) is ((4pi^2+4)^{3/2}-8)/3

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integral from 0 to 2pi of sqrt(θ^4+4θ^2)

\int_{0}^{2 π} θ^{4} + 4 θ^{2} d θ

\frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

Decimal

92.89623 \dots

Solution steps

\int_{0}^{2 π} θ^{4} + 4 θ^{2} d θ

Simplify

θ^{4} + 4 θ^{2} : θ θ^{2} + 4

= \int_{0}^{2 π} θ θ^{2} + 4 d θ

Apply u-substitution

: \int_{2}^{4 π^{2} + 4} u^{2} d u

= \int_{2}^{4 π^{2} + 4} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{2}^{4 π^{2} + 4}

= [\frac{u ^{3}}{3}]_{2}^{4 π^{2} + 4}

Compute the boundaries:

\frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}}}{3} - \frac{8}{3}

= \frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}}}{3} - \frac{8}{3}

Simplify

= \frac{( 4 π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

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What is the integral from 0 to 2pi of sqrt(θ^4+4θ^2) ?
The integral from 0 to 2pi of sqrt(θ^4+4θ^2) is ((4pi^2+4)^{3/2}-8)/3