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

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

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

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

\frac{14}{9}

Decimal

1.55555 \dots

Solution steps

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

Apply u-substitution

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

= \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}

Compute the boundaries:

\frac{14}{3}

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

Simplify

= \frac{14}{9}

y = - 4 x - 3 y = - 4 x - 3

\int_{0}^{3} π (\frac{2}{3} x)^{2} d x

- 3 sin (3 t)

\frac{d}{d x} (sin^{- 2} (x))

f (x) = \frac{1 - 3 x}{4 + x}

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