Let’s evaluate the two integrals step by step.

(a) $\int_0^{\sqrt{3}/2} \sqrt{1 - x^2} \, dx$

This integral resembles the formula for the area of a circular segment. Specifically, $\sqrt{1 - x^2}$ corresponds to the upper semicircle of a unit circle. The integral can be directly computed using a trigonometric substitution:

1. Substitution: Let $x = \sin \theta$, so $dx = \cos \theta \, d\theta$ and $\sqrt{1 - x^2} = \cos \theta$.

   The limits transform as: $x = 0 \implies \theta = 0, \quad x = \sqrt{3}/2 \implies \theta = \pi/3$

2. Rewrite the integral: $\int_0^{\sqrt{3}/2} \sqrt{1 - x^2} \, dx = \int_0^{\pi/3} \cos^2 \theta \, d\theta$

3. Simplify using a trigonometric identity: $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$

   Substitute: $\int_0^{\pi/3} \cos^2 \theta \, d\theta = \int_0^{\pi/3} \frac{1 + \cos(2\theta)}{2} \, d\theta$

   Break into two integrals: $= \frac{1}{2} \int_0^{\pi/3} 1 \, d\theta + \frac{1}{2} \int_0^{\pi/3} \cos(2\theta) \, d\theta$

4. Evaluate each term:

   • First term: $\frac{1}{2} \int_0^{\pi/3} 1 \, d\theta = \frac{1}{2} \cdot \left[\theta\right]_0^{\pi/3} = \frac{1}{2} \cdot \frac{\pi}{3} = \frac{\pi}{6}$

   • Second term: $\frac{1}{2} \int_0^{\pi/3} \cos(2\theta) \, d\theta = \frac{1}{2} \cdot \frac{\sin(2\theta)}{2} \Big|_0^{\pi/3} = \frac{1}{4} \left[\sin\left(\frac{2\pi}{3}\right) - \sin(0)\right]$

     Since $\sin\left(\frac{2\pi}{3}\right) = \sqrt{3}/2$, we get: $= \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$

5. Combine results: $\int_0^{\sqrt{3}/2} \sqrt{1 - x^2} \, dx = \frac{\pi}{6} + \frac{\sqrt{3}}{8}$

(b) $\int_0^1 \sqrt{x} (1 + 6x^2) \, dx$

1. Simplify: Rewrite $\sqrt{x} = x^{1/2}$: $\int_0^1 \sqrt{x} (1 + 6x^2) \, dx = \int_0^1 x^{1/2} \, dx + 6 \int_0^1 x^{1/2} x^2 \, dx$

   Combine powers of $x$: $= \int_0^1 x^{1/2} \, dx + 6 \int_0^1 x^{5/2} \, dx$

2. Evaluate each term:

   • First term: $\int_0^1 x^{1/2} \, dx = \left[\frac{x^{3/2}}{3/2}\right]_0^1 = \frac{2}{3} x^{3/2} \Big|_0^1 = \frac{2}{3}$

   • Second term: $\int_0^1 x^{5/2} \, dx = \left[\frac{x^{7/2}}{7/2}\right]_0^1 = \frac{2}{7} x^{7/2} \Big|_0^1 = \frac{2}{7}$

3. Combine results: $\int_0^1 \sqrt{x} (1 + 6x^2) \, dx = \frac{2}{3} + 6 \cdot \frac{2}{7} = \frac{2}{3} + \frac{12}{7}$

   Find a common denominator: $\frac{2}{3} = \frac{14}{21}, \quad \frac{12}{7} = \frac{36}{21}$

   Add: $\frac{14}{21} + \frac{36}{21} = \frac{50}{21}$

Final Answers:

(a) $\frac{\pi}{6} + \frac{\sqrt{3}}{8}$

(b) $\frac{50}{21}$

Would you like further clarification or a visual representation of these computations?
Here are 5 related questions to deepen understanding:

1. How does trigonometric substitution simplify integrals involving radicals?
2. What is the geometric interpretation of the integral in part (a)?
3. Can you derive a formula for the area under a semicircle using definite integrals?
4. How does the power rule apply in fractional powers for part (b)?
5. Why is it important to properly combine fractions when solving integrals?

Tip: Always check the limits of substitution carefully, especially when transforming definite integrals.