What is the integral from 0 to 1 of xsqrt(4-3x) ?

The integral from 0 to 1 of xsqrt(4-3x) is 94/135

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integral from 0 to 1 of xsqrt(4-3x)

\int_{0}^{1} x 4 - 3 x d x

\frac{94}{135}

Decimal

0.69629 \dots

Solution steps

\int_{0}^{1} x 4 - 3 x d x

Apply u-substitution

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

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

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

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

Expand

u (u - 4) : u^{\frac{3}{2}} - 4 u

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

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

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

\int_{1}^{4} 4 u d u = \frac{56}{3}

= - \frac{1}{9} (\frac{62}{5} - \frac{56}{3})

Simplify

- \frac{1}{9} (\frac{62}{5} - \frac{56}{3}) : \frac{94}{135}

= \frac{94}{135}

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What is the integral from 0 to 1 of xsqrt(4-3x) ?
The integral from 0 to 1 of xsqrt(4-3x) is 94/135