# How do you find the definite integral of #int sqrt(x) dx# from #[0,3]#?

Been quite a while since I did any Integration so hopefully I am correct!

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To find the definite integral of (\int \sqrt{x} , dx) from (0) to (3), use the following steps:

- Integrate (\sqrt{x}) with respect to (x) to find the antiderivative.
- Evaluate the antiderivative at the upper limit of integration (in this case, (3)).
- Subtract the value of the antiderivative at the lower limit of integration (in this case, (0)).

The antiderivative of (\sqrt{x}) is (\frac{2}{3} x^{\frac{3}{2}} + C), where (C) is the constant of integration.

Evaluate the antiderivative at (x = 3): [ \frac{2}{3} \cdot 3^{\frac{3}{2}} = \frac{2}{3} \cdot 3^{\frac{3}{2}} = \frac{2}{3} \cdot 3 \cdot \sqrt{3} = 2\sqrt{3} ]

Evaluate the antiderivative at (x = 0): [ \frac{2}{3} \cdot 0^{\frac{3}{2}} = 0 ]

Subtract the lower limit value from the upper limit value: [ 2\sqrt{3} - 0 = 2\sqrt{3} ]

Therefore, the definite integral of (\int \sqrt{x} , dx) from (0) to (3) is (2\sqrt{3}).

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