What is the integral from 0 to 1 of sqrt(9t+4) ?

The integral from 0 to 1 of sqrt(9t+4) is (26sqrt(13)-16)/(27)

Popular >

integral from 0 to 1 of sqrt(9t+4)

\int_{0}^{1} 9 t + 4 d t

\frac{26 13 - 16}{27}

Decimal

2.87941 \dots

Solution steps

\int_{0}^{1} 9 t + 4 d t

Apply u-substitution

: \int_{4}^{13} \frac{u}{9} d u

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

Apply the Power Rule

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

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

Compute the boundaries:

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

= \frac{1}{9} (\frac{26 13}{3} - \frac{16}{3})

Simplify

= \frac{26 13 - 16}{27}

\int_{1}^{2} (3 x^{2} - 2 x + 4) d x

\int_{- 2}^{5} x d x

\int_{0}^{2} (x - 2) (x - 4) d x

\int_{0}^{π} cos (d) x d x

\int_{0}^{2} π y^{4} d y

What is the integral from 0 to 1 of sqrt(9t+4) ?
The integral from 0 to 1 of sqrt(9t+4) is (26sqrt(13)-16)/(27)