What is the integral from-1 to 0 of sqrt(3u+4) ?

The integral from-1 to 0 of sqrt(3u+4) is 14/9

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integral from-1 to 0 of sqrt(3u+4)

\int_{- 1}^{0} 3 u + 4 d u

\frac{14}{9}

Decimal Notation

1.55555 \dots

Solution steps

\int_{- 1}^{0} 3 u + 4 d u

Apply u-substitution

: \int_{1}^{4} \frac{v}{3} d v

= \int_{1}^{4} \frac{v}{3} d v

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} v d v

Apply the Power Rule

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

= \frac{1}{3} [\frac{2}{3} v^{\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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What is the integral from-1 to 0 of sqrt(3u+4) ?
The integral from-1 to 0 of sqrt(3u+4) is 14/9