1. **Problem:** Evaluate the integral $$\int_0^3 \sqrt{y+1} \, dy$$. 2. **Formula and rules:** Use the substitution for integrals of the form $$\int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2} + C$$. 3. **Intermediate work:** - Let $$u = y + 1$$, then $$du = dy$$. - Change limits: when $$y=0$$, $$u=1$$; when $$y=3$$, $$u=4$$. - The integral becomes $$\int_1^4 u^{1/2} \, du$$. - Integrate: $$\frac{2}{3} u^{3/2} \Big|_1^4 = \frac{2}{3} (4^{3/2} - 1^{3/2})$$. - Calculate powers: $$4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$$. - So, $$\frac{2}{3} (8 - 1) = \frac{2}{3} \times 7 = \frac{14}{3}$$. 4. **Answer:** $$\int_0^3 \sqrt{y+1} \, dy = \frac{14}{3}$$.