What is the integral from 2 to 6 of sqrt(4x+1) ?

The integral from 2 to 6 of sqrt(4x+1) is 49/3

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integral from 2 to 6 of sqrt(4x+1)

\int_{2}^{6} 4 x + 1 d x

\frac{49}{3}

Decimal

16.33333 \dots

Solution steps

\int_{2}^{6} 4 x + 1 d x

Apply u-substitution

: \int_{9}^{25} \frac{u}{4} d u

= \int_{9}^{25} \frac{u}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{9}^{25} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{25}

Compute the boundaries:

\frac{196}{3}

= \frac{1}{4} \cdot \frac{196}{3}

Simplify

= \frac{49}{3}

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\int_{3}^{\infty} e^{- s t} d t

\int_{0}^{9} (81 - x^{2}) d x

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What is the integral from 2 to 6 of sqrt(4x+1) ?
The integral from 2 to 6 of sqrt(4x+1) is 49/3