What is the integral from 0 to 6pi of θsqrt(θ^2+4) ?

The integral from 0 to 6pi of θsqrt(θ^2+4) is ((36pi^2+4)^{3/2}-8)/3

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integral from 0 to 6pi of θsqrt(θ^2+4)

\int_{0}^{6 π} θ θ^{2} + 4 d θ

\frac{( 36 π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

Decimal Notation

2267.59027 \dots

Solution steps

\int_{0}^{6 π} θ θ^{2} + 4 d θ

Apply u-substitution

: \int_{2}^{36 π^{2} + 4} u^{2} d u

= \int_{2}^{36 π^{2} + 4} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{2}^{36 π^{2} + 4}

= [\frac{u ^{3}}{3}]_{2}^{36 π^{2} + 4}

Compute the boundaries:

\frac{( 36 π ^{2} + 4 ) ^{\frac{3}{2}}}{3} - \frac{8}{3}

= \frac{( 36 π ^{2} + 4 ) ^{\frac{3}{2}}}{3} - \frac{8}{3}

Simplify

= \frac{( 36 π ^{2} + 4 ) ^{\frac{3}{2}} - 8}{3}

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What is the integral from 0 to 6pi of θsqrt(θ^2+4) ?
The integral from 0 to 6pi of θsqrt(θ^2+4) is ((36pi^2+4)^{3/2}-8)/3