Evaluate the following integrals.



**(a)**

√

3

**/** 2

∫

0

5

√

1 − *x*2

 *dx*

**(b)**

1

∫

0

√

*x*

 (1 + 6*x*2 ) *dx*



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.

Evaluate the following integrals:

(a) ∫₀^(√3/2) √(1 - x²) dx
(b) ∫₀¹ √x (1 + 6x²) dx

Explore the detailed solution for two integrals: (a) ∫₀^(√3/2) √(1 - x²) dx using trigonometric substitution, and (b) ∫₀¹ √x (1 + 6x²) dx using the power rule. Perfect for college-level calculus students.

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