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

The integral from-1 to 0 of sqrt(1-3x) is 14/9

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

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

\frac{14}{9}

Decimal Notation

1.55555 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{1} - \frac{u}{3} d u

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

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

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

= - (- \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4})

Simplify

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{1}{3} \cdot \frac{14}{3}

Simplify

= \frac{14}{9}

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\int_{0}^{2 π} - 2 sin^{2} (x) d x

What is the integral from-1 to 0 of sqrt(1-3x) ?
The integral from-1 to 0 of sqrt(1-3x) is 14/9