What is the integral from 0 to 3 of 2pix(9-x^2) ?

The integral from 0 to 3 of 2pix(9-x^2) is (81pi)/2

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integral from 0 to 3 of 2pix(9-x^2)

\int_{0}^{3} 2 π x (9 - x^{2}) d x

\frac{81 π}{2}

Decimal Notation

127.23450 \dots

Solution steps

\int_{0}^{3} 2 π x (9 - x^{2}) 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}^{3} x (9 - x^{2}) d x

Expand

x (9 - x^{2}) : 9 x - x^{3}

= 2 π \cdot \int_{0}^{3} 9 x - x^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 π (\int_{0}^{3} 9 x d x - \int_{0}^{3} x^{3} d x)

\int_{0}^{3} 9 x d x = \frac{81}{2}

\int_{0}^{3} x^{3} d x = \frac{81}{4}

= 2 π (\frac{81}{2} - \frac{81}{4})

Simplify

2 π (\frac{81}{2} - \frac{81}{4}) : \frac{81 π}{2}

= \frac{81 π}{2}

\frac{d}{d x} (5 x^{l n (5 x)})

y^{^{'}} + 2 x y = \frac{x}{y ^{2}}

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{8} d x

y = (t \cdot sin (t))^{2} + t \cdot sin (t)

sec^{2} (2 x) - tan^{2} (2 x)

What is the integral from 0 to 3 of 2pix(9-x^2) ?
The integral from 0 to 3 of 2pix(9-x^2) is (81pi)/2