What is the integral from 1 to 2 of sqrt(1+9x) ?

The integral from 1 to 2 of sqrt(1+9x) is (38sqrt(19)-20sqrt(10))/(27)

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integral from 1 to 2 of sqrt(1+9x)

\int_{1}^{2} 1 + 9 x d x

\frac{38 19 - 20 10}{27}

Decimal

3.79231 \dots

Solution steps

\int_{1}^{2} 1 + 9 x d x

Apply u-substitution

= \int_{10}^{19} \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_{10}^{19} u d u

Apply the Power Rule

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{10}^{19}

Compute the boundaries:

\frac{38 19 - 20 10}{3}

= \frac{1}{9} \cdot \frac{38 19 - 20 10}{3}

Simplify

= \frac{38 19 - 20 10}{27}

What is the integral from 1 to 2 of sqrt(1+9x) ?
The integral from 1 to 2 of sqrt(1+9x) is (38sqrt(19)-20sqrt(10))/(27)