What is the integral from 0 to 2 of sqrt(1+4x) ?

The integral from 0 to 2 of sqrt(1+4x) is 13/3

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integral from 0 to 2 of sqrt(1+4x)

\int_{0}^{2} 1 + 4 x d x

\frac{13}{3}

Decimal

4.33333 \dots

Solution steps

\int_{0}^{2} 1 + 4 x d x

Apply u-substitution

= \int_{1}^{9} \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_{1}^{9} u d u

Apply the Power Rule

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

Compute the boundaries:

\frac{52}{3}

= \frac{1}{4} \cdot \frac{52}{3}

Simplify

= \frac{13}{3}

What is the integral from 0 to 2 of sqrt(1+4x) ?
The integral from 0 to 2 of sqrt(1+4x) is 13/3