What is the integral from 0 to 3 of sqrt((3t+4)) ?

The integral from 0 to 3 of sqrt((3t+4)) is (26sqrt(13)-16)/9

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integral from 0 to 3 of sqrt((3t+4))

\int_{0}^{3} (3 t + 4) d t

\frac{26 13 - 16}{9}

Decimal Notation

8.63825 \dots

Solution steps

\int_{0}^{3} 3 t + 4 d t

Apply u-substitution

: \int_{4}^{13} \frac{u}{3} d u

= \int_{4}^{13} \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_{4}^{13} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

Compute the boundaries:

\frac{26 13}{3} - \frac{16}{3}

= \frac{1}{3} (\frac{26 13}{3} - \frac{16}{3})

Simplify

= \frac{26 13 - 16}{9}

\int_{- 1}^{1} ln ∣ x ∣ d x

y = \frac{13}{x ^{2} + 5 x + 6}, y = 0, x = 0, x = 1

12 y^{^{''}} - 12 y^{^{'}} + 78 y = 0

\int 5 e^{- 0.5 x} d x

\frac{d}{d x} (65 - 55 cos (\frac{π x}{15}))

What is the integral from 0 to 3 of sqrt((3t+4)) ?
The integral from 0 to 3 of sqrt((3t+4)) is (26sqrt(13)-16)/9