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

The integral from 0 to 1 of sqrt(x+9) is (20sqrt(10))/3-18

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

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

\frac{20 10}{3} - 18

Decimal Notation

3.08185 \dots

Solution steps

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

Apply u-substitution

: \int_{9}^{10} u d u

= \int_{9}^{10} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{10}

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

Compute the boundaries:

\frac{20 10}{3} - 18

= \frac{20 10}{3} - 18

\frac{\partial}{\partial x} (e^{- r x})

\frac{3}{x + 7}

\frac{d}{d x} (ln (ln (2 x + 2 y)))

\int 0.2 x^{2} + 1 d x

What is the integral from 0 to 1 of sqrt(x+9) ?
The integral from 0 to 1 of sqrt(x+9) is (20sqrt(10))/3-18