What is the integral from 0 to 3 of (3x(8-x))/(256) ?

The integral from 0 to 3 of (3x(8-x))/(256) is 81/256

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integral from 0 to 3 of (3x(8-x))/(256)

\int_{0}^{3} \frac{3 x ( 8 - x )}{256} d x

\frac{81}{256}

Decimal

0.31640625

Solution steps

\int_{0}^{3} \frac{3 x ( 8 - x )}{256} d x

Take the constant out:

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

= \frac{3}{256} \cdot \int_{0}^{3} x (8 - x) d x

Expand

x (8 - x) : 8 x - x^{2}

= \frac{3}{256} \cdot \int_{0}^{3} 8 x - x^{2} 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

= \frac{3}{256} (\int_{0}^{3} 8 x d x - \int_{0}^{3} x^{2} d x)

\int_{0}^{3} 8 x d x = 36

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

= \frac{3}{256} (36 - 9)

Simplify

\frac{3}{256} (36 - 9) : \frac{81}{256}

= \frac{81}{256}

\int_{1}^{\infty} \frac{1}{2 x + 3} d x

\int_{0}^{\frac{π}{2}} \frac{1}{3 + cos ( 2 a )} d a

\int_{- 1}^{3} (- 3 x^{2} + 8 x + 5) d x

\int_{- 1}^{1} 8^{x} d x

\int_{0}^{6} x^{2} - 6 x + 5 d x

What is the integral from 0 to 3 of (3x(8-x))/(256) ?
The integral from 0 to 3 of (3x(8-x))/(256) is 81/256