What is the integral from 0 to 6 of 2pixsqrt(x^2+9) ?

The integral from 0 to 6 of 2pixsqrt(x^2+9) is 2pi(45sqrt(5)-9)

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integral from 0 to 6 of 2pixsqrt(x^2+9)

\int_{0}^{6} 2 π x x^{2} + 9 d x

2 π (455 - 9)

Decimal

575.68465 \dots

Solution steps

\int_{0}^{6} 2 π x x^{2} + 9 d x

Take the constant out:

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

= 2 π \cdot \int_{0}^{6} x x^{2} + 9 d x

Apply u-substitution

: \int_{3}^{35} u^{2} d u

= 2 π \cdot \int_{3}^{35} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{3}^{35}

= 2 π [\frac{u ^{3}}{3}]_{3}^{35}

Compute the boundaries:

455 - 9

= 2 π (455 - 9)

x^{3} + 8

\frac{d}{d x} (\frac{3}{2} x^{\frac{1}{2}} - \frac{1}{3} x^{\frac{2}{3}})

t \to 0 lim (\frac{2}{1 + t})

\frac{\partial}{\partial x} (e^{x} y)

\int_{0}^{8} (1 + 3 x) d x

What is the integral from 0 to 6 of 2pixsqrt(x^2+9) ?
The integral from 0 to 6 of 2pixsqrt(x^2+9) is 2pi(45sqrt(5)-9)