What is the integral from-1 to 0 of 3xsqrt(1-3x) ?

The integral from-1 to 0 of 3xsqrt(1-3x) is -116/45

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integral from-1 to 0 of 3xsqrt(1-3x)

\int_{- 1}^{0} 3 x 1 - 3 x d x

- \frac{116}{45}

Solution steps

\int_{- 1}^{0} 3 x 1 - 3 x d x

Take the constant out:

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

= 3 \cdot \int_{- 1}^{0} x 1 - 3 x d x

Apply u-substitution

: \int_{4}^{1} \frac{u ( u - 1 )}{9} d u

= 3 \cdot \int_{4}^{1} \frac{u ( u - 1 )}{9} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 3 (- \int_{1}^{4} \frac{u ( u - 1 )}{9} d u)

Take the constant out:

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

= 3 (- \frac{1}{9} \cdot \int_{1}^{4} u (u - 1) d u)

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= 3 (- \frac{1}{9} \cdot \int_{1}^{4} u^{\frac{3}{2}} - u d u)

Apply the Sum Rule:

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

= 3 (- \frac{1}{9} (\int_{1}^{4} u^{\frac{3}{2}} d u - \int_{1}^{4} u d u))

\int_{1}^{4} u^{\frac{3}{2}} d u = \frac{62}{5}

\int_{1}^{4} u d u = \frac{14}{3}

= 3 (- \frac{1}{9} (\frac{62}{5} - \frac{14}{3}))

Simplify

3 (- \frac{1}{9} (\frac{62}{5} - \frac{14}{3})) : - \frac{116}{45}

= - \frac{116}{45}

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What is the integral from-1 to 0 of 3xsqrt(1-3x) ?
The integral from-1 to 0 of 3xsqrt(1-3x) is -116/45