What is the integral from-3 to 6 of sqrt(3+x) ?

The integral from-3 to 6 of sqrt(3+x) is 18

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integral from-3 to 6 of sqrt(3+x)

\int_{- 3}^{6} 3 + x d x

18

Solution steps

\int_{- 3}^{6} 3 + x d x

Apply u-substitution

: \int_{0}^{9} u d u

= \int_{0}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

Compute the boundaries:

18

= 18

\frac{\partial}{\partial x} (e^{9 z})

x \to 0 lim (1 + x^{2})

\int (e^{y}) d y

1 + 3 x, 1 \leq x \leq 3

\int_{0}^{16} \frac{1}{2} x d x

What is the integral from-3 to 6 of sqrt(3+x) ?
The integral from-3 to 6 of sqrt(3+x) is 18