Question: The area of the quarter circle shown shaded on the right is given by $\int_0^5 \sqrt{25 - x^2} \, dx$. Use the substitution $x = 5 \sin u$ to evaluate this definite integral exactly and show that your answer is consistent with the area of a circle of radius $r$ being $\pi r^2$.
1. **State the problem:** We want to evaluate the integral $$\int_0^5 \sqrt{25 - x^2} \, dx$$ using the substitution $x = 5 \sin u$ and verify the result matches the quarter circle area formula.
2. **Recall the formula for the area of a circle:** The area of a circle with radius $r$ is $$\pi r^2$$. A quarter circle area is therefore $$\frac{\pi r^2}{4}$$.
3. **Substitution:** Let $$x = 5 \sin u$$. Then $$dx = 5 \cos u \, du$$.
4. **Change the limits:** When $x=0$, $$0 = 5 \sin u \implies \sin u = 0 \implies u=0$$.
When $x=5$, $$5 = 5 \sin u \implies \sin u = 1 \implies u = \frac{\pi}{2}$$.
5. **Rewrite the integral:**
$$\int_0^5 \sqrt{25 - x^2} \, dx = \int_0^{\pi/2} \sqrt{25 - (5 \sin u)^2} \cdot 5 \cos u \, du$$
6. **Simplify the integrand:**
$$\sqrt{25 - 25 \sin^2 u} = \sqrt{25(1 - \sin^2 u)} = \sqrt{25 \cos^2 u} = 5 \cos u$$
7. **Substitute back:**
$$\int_0^{\pi/2} 5 \cos u \cdot 5 \cos u \, du = \int_0^{\pi/2} 25 \cos^2 u \, du$$
8. **Use the identity:**
$$\cos^2 u = \frac{1 + \cos 2u}{2}$$
9. **Integral becomes:**
$$25 \int_0^{\pi/2} \frac{1 + \cos 2u}{2} \, du = \frac{25}{2} \int_0^{\pi/2} (1 + \cos 2u) \, du$$
10. **Integrate term by term:**
$$\frac{25}{2} \left[ u + \frac{\sin 2u}{2} \right]_0^{\pi/2} = \frac{25}{2} \left( \frac{\pi}{2} + 0 - 0 \right) = \frac{25}{2} \cdot \frac{\pi}{2} = \frac{25 \pi}{4}$$
11. **Interpretation:** The integral equals $$\frac{25 \pi}{4}$$ which is the area of a quarter circle with radius $5$ since $$\frac{\pi r^2}{4} = \frac{\pi \cdot 5^2}{4} = \frac{25 \pi}{4}$$.
**Final answer:** $$\int_0^5 \sqrt{25 - x^2} \, dx = \frac{25 \pi}{4}$$ which matches the quarter circle area formula.