What is the integral from 0 to 7/3 of sqrt(9+3x) ?

The integral from 0 to 7/3 of sqrt(9+3x) is 74/9

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integral from 0 to 7/3 of sqrt(9+3x)

\int_{0}^{\frac{7}{3}} 9 + 3 x d x

\frac{74}{9}

Decimal Notation

8.22222 \dots

Solution steps

\int_{0}^{\frac{7}{3}} 9 + 3 x d x

Apply u-substitution

: \int_{9}^{16} \frac{u}{3} d u

= \int_{9}^{16} \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{9}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{16}

Compute the boundaries:

\frac{74}{3}

= \frac{1}{3} \cdot \frac{74}{3}

Simplify

= \frac{74}{9}

\frac{\partial}{\partial x} (lo g_{e} (x^{2} + y^{2}))

\int \frac{( 5 t - t )}{3 t ^{2} - 9} d t

\frac{\partial}{\partial x} (x^{4} - 3 x^{2} y^{2} + y^{4})

x \to - 2 lim (4^{2} - 3)

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 0, y (\frac{π}{2}) = 1

What is the integral from 0 to 7/3 of sqrt(9+3x) ?
The integral from 0 to 7/3 of sqrt(9+3x) is 74/9