What is the integral from 0 to 2pi of sqrt(400+9t) ?

The integral from 0 to 2pi of sqrt(400+9t) is (2(400+18pi)^{3/2}-16000)/(27)

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integral from 0 to 2pi of sqrt(400+9t)

\int_{0}^{2 π} 400 + 9 t d t

\frac{2 ( 400 + 18 π ) ^{\frac{3}{2}} - 16000}{27}

Decimal

130.00556 \dots

Solution steps

\int_{0}^{2 π} 400 + 9 t d t

Apply u-substitution

= \int_{400}^{400 + 18 π} \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{400}^{400 + 18 π} u d u

Apply the Power Rule

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{400}^{400 + 18 π}

Compute the boundaries:

\frac{2}{3} (400 + 18 π)^{\frac{3}{2}} - \frac{16000}{3}

= \frac{1}{9} (\frac{2}{3} (400 + 18 π)^{\frac{3}{2}} - \frac{16000}{3})

Simplify

= \frac{2 ( 400 + 18 π ) ^{\frac{3}{2}} - 16000}{27}

What is the integral from 0 to 2pi of sqrt(400+9t) ?
The integral from 0 to 2pi of sqrt(400+9t) is (2(400+18pi)^{3/2}-16000)/(27)