What would you like to understand?

To evaluate $\int_2^3 \sqrt{x} dx$, first find the antiderivative $\frac{2}{3}x^{3/2}$. Then, apply the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from its value at the upper bound: $\left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}(3)^{3/2} - \frac{2}{3}(2)^{3/2}$.

Conditions: The integral is $\int_2^3 \sqrt{x} dx$.; The antiderivative is $\frac{2}{3}x^{3/2}$.; The bounds are 2 and 3.

To evaluate $\int_2^3 \sqrt{x} dx$, first find the antiderivative $\frac{2}{3}x^{3/2}$. Then, apply the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from its value at the upper bound: $\left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}(3)^{3/2} - \frac{2}{3}(2)^{3/2}$.

Conditions: The integral is $\int_2^3 \sqrt{x} dx$.; The antiderivative is $\frac{2}{3}x^{3/2}$.; The bounds are 2 and 3.