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

The integral from 0 to 4 of sqrt(1+9x) is (74sqrt(37)-2)/(27)

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

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

\frac{74 37 - 2}{27}

Decimal Notation

16.59720 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{37} \frac{u}{9} d u

= \int_{1}^{37} \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_{1}^{37} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

Compute the boundaries:

\frac{74 37}{3} - \frac{2}{3}

= \frac{1}{9} (\frac{74 37}{3} - \frac{2}{3})

Simplify

= \frac{74 37 - 2}{27}

ln (1 - 3 y)

\int y cos (y^{2}) d y

\frac{d}{d x} (\frac{x \cdot sin ( x )}{1 + x ^{2}})

\int_{0}^{π} 1 - cos^{2} (x) d x

\frac{d}{d x} (8 e^{π x} + 2 e^{7 x})

What is the integral from 0 to 4 of sqrt(1+9x) ?
The integral from 0 to 4 of sqrt(1+9x) is (74sqrt(37)-2)/(27)