Certainly! However, the function to be considered is not specified in your image. Typically, a Fundamental Theorem of Calculus worksheet might use a function like \( f(x) = \sqrt{R^2 - x^2} \), which describes the upper half of a circle of radius \( R \) centered at the origin. This is a common example for such problems. Let's proceed with **\( f(x) = \sqrt{R^2 - x^2} \)** for this worksheet, assuming \( R > 0 \). --- ### 1. **Consider the function \( f(x) = \sqrt{R^2 - x^2} \). Find \( f'(x) \) and show that.** **Solution:** Given: \( f(x) = \sqrt{R^2 - x^2} \) Let’s use the chain rule: \[ f'(x) = \frac{d}{dx} \left( (R^2 - x^2)^{1/2} \right) = \frac{1}{2} (R^2 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{R^2 - x^2}} \] **Final answer:** \[ f'(x) = \frac{-x}{\sqrt{R^2 - x^2}} \] --- ### 2. **Find the following indefinite integral.** \[ \int \sqrt{R^2 - x^2} \, dx \] **Solution:** This is a standard integral. Use the substitution \( x = R \sin \theta \), \( dx = R \cos \theta d\theta \): \[ \int \sqrt{R^2 - x^2} dx = \int \sqrt{R^2 - R^2 \sin^2\theta} \cdot R\cos\theta d\theta = \int \sqrt{R^2(1 - \sin^2\theta)} \cdot R\cos\theta d\theta = \int R \cos\theta \cdot R \cos\theta d\theta = R^2 \int \cos^2\theta d\theta \] Using the double angle identity: \[ \cos^2\theta = \frac{1 + \cos(2\theta)}{2} \] So, \[ R^2 \int \cos^2\theta d\theta = R^2 \int \frac{1 + \cos(2\theta)}{2} d\theta = \frac{R^2}{2} \int (1 + \cos(2\theta)) d\theta = \frac{R^2}{2} \left( \theta + \frac{1}{2}\sin(2\theta) \right) + C \] Transform back to \( x \): - \( x = R\sin\theta \implies \theta = \arcsin(x/R) \) - \( \sin(2\theta) = 2\sin\theta\cos\theta = 2(x/R)\left(\sqrt{1 - x^2/R^2}\right) = 2x\sqrt{R^2-x^2}/R^2 \) So, \[ \int \sqrt{R^2 - x^2} dx = \frac{R^2}{2} \arcsin\left(\frac{x}{R}\right) + \frac{x}{2}\sqrt{R^2 - x^2} + C \] --- ### 3. **What is the domain of \( f(x) \)?** **Solution:** Since \( f(x) = \sqrt{R^2 - x^2} \), the expression under the square root must be non-negative: \[ R^2 - x^2 \geq 0 \implies -R \leq x \leq R \] **Domain:** \( [-R, R] \) --- ### 4. **Draw the graph of the function on the domain and shade the area calculated by the definite integral.** **Solution:** - The graph of \( f(x) = \sqrt{R^2 - x^2} \) on \( [-R, R] \) is the upper half of a circle with radius \( R \). - The area under this curve from \( -R \) to \( R \) is the area of the upper semicircle. *(You would sketch a semicircle above the x-axis from \( x = -R \) to \( x = R \), shading the area under the curve.)* --- ### 5. **Using the Fundamental Theorem of Calculus, calculate the definite integral:** \[ \int_{-R}^{R} \sqrt{R^2 - x^2} dx \] **Solution:** Using the result from question 2: \[ F(x) = \frac{R^2}{2} \arcsin\left(\frac{x}{R}\right) + \frac{x}{2} \sqrt{R^2 - x^2} \] Evaluate at \( x = R \) and \( x = -R \): - At \( x = R \): - \( \arcsin(1) = \frac{\pi}{2} \) - \( \sqrt{R^2 - R^2} = 0 \) - At \( x = -R \): - \( \arcsin(-1) = -\frac{\pi}{2} \) - \( \sqrt{R^2 - R^2} = 0 \) So, \[ F(R) = \frac{R^2}{2} \cdot \frac{\pi}{2} + 0 = \frac{\pi R^2}{4} \] \[ F(-R) = \frac{R^2}{2} \cdot \left(-\frac{\pi}{2}\right) + 0 = -\frac{\pi R^2}{4} \] So, \[ \int_{-R}^{R} \sqrt{R^2 - x^2} dx = F(R) - F(-R) = \frac{\pi R^2}{4} - \left(-\frac{\pi R^2}{4}\right) = \frac{\pi R^2}{2} \] --- ### 6. **Explain the geometric significance of the answer in question 5.** **Solution:** The value \( \frac{\pi R^2}{2} \) is **half the area of a circle** with radius \( R \). This makes sense because the graph of \( f(x) = \sqrt{R^2 - x^2} \) from \( x = -R \) to \( x = R \) traces out the upper half of a circle of radius \( R \). Thus, the definite integral gives the area under the semicircle, which is exactly half the area of the full circle. Full circle area: \( \pi R^2 \) Upper semicircle area: \( \frac{1}{2}\pi R^2 = \frac{\pi R^2}{2} \) --- ## **Summary Table:** | Step | Solution/Explanation | |------|---------------------| | 1 | \( f'(x) = \frac{-x}{\sqrt{R^2 - x^2}} \) | | 2 | \( \int \sqrt{R^2 - x^2} dx = \frac{R^2}{2} \arcsin\left(\frac{x}{R}\right) + \frac{x}{2}\sqrt{R^2-x^2} + C \) | | 3 | Domain: \( [-R, R] \) | | 4 | Upper semicircle on \( [-R, R] \), area under curve shaded | | 5 | \( \int_{-R}^{R} \sqrt{R^2 - x^2} dx = \frac{\pi R^2}{2} \) | | 6 | Geometric significance: This is the area of a semicircle of radius \( R \). | If your function was different, please specify and I will adjust the solution!

Certainly! Let's proceed step-by-step with detailed calculations and explanations for each part, assuming the function is: \[ f(x) = \sqrt{R^2 - x^2} \] This is a common function used in calculus related to the circle's upper semicircle. --- ### **Part 1: Find \(f'(x)\) and show that** **Given:** \[ f(x) = \sqrt{R^2 - x^2} \] **Step-by-step:** - Rewrite \(f(x)\) as: \[ f(x) = (R^2 - x^2)^{1/2} \] - Use the chain rule for derivatives: \[ f'(x) = \frac{d}{dx} (R^2 - x^2)^{1/2} = \frac{1}{2} (R^2 - x^2)^{-1/2} \times \frac{d}{dx} (R^2 - x^2) \] - Derivative of the inner function: \[ \frac{d}{dx} (R^2 - x^2) = -2x \] - Therefore, \[ f'(x) = \frac{1}{2} (R^2 - x^2)^{-1/2} \times (-2x) = -x (R^2 - x^2)^{-1/2} \] - Rewrite: \[ f'(x) = \frac{-x}{\sqrt{R^2 - x^2}} \] **Final answer:** \[ \boxed{ f'(x) = \frac{-x}{\sqrt{R^2 - x^2}} } \] --- ### **Part 2: Find the indefinite integral** \[ I = \int \sqrt{R^2 - x^2} \, dx \] **Step-by-step solution:** **Method:** Use substitution based on a trigonometric substitution \(x = R \sin \theta\). - **Substitution:** \[ x = R \sin \theta \implies dx = R \cos \theta \, d\theta \] - **Express the integrand:** \[ \sqrt{R^2 - x^2} = \sqrt{R^2 - R^2 \sin^2 \theta} = \sqrt{R^2 (1 - \sin^2 \theta)} = R \cos \theta \] - **Rewrite the integral:** \[ I = \int R \cos \theta \times R \cos \theta \, d\theta = R^2 \int \cos^2 \theta \, d\theta \] - **Integral of \(\cos^2 \theta\):** Recall the identity: \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \] - **Integrate:** \[ I = R^2 \int \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{R^2}{2} \int (1 + \cos 2\theta) \, d\theta \] \[ = \frac{R^2}{2} \left( \int 1 \, d\theta + \int \cos 2\theta \, d\theta \right) \] - **Evaluate:** \[ = \frac{R^2}{2} \left( \theta + \frac{1}{2} \sin 2\theta \right) + C \] - **Back-substitute:** Recall: \[ x = R \sin \theta \implies \theta = \arcsin \left( \frac{x}{R} \right) \] and: \[ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \frac{x}{R} \times \frac{\sqrt{R^2 - x^2}}{R} = \frac{2x \sqrt{R^2 - x^2}}{R^2} \] - **Final expression:** \[ I = \frac{R^2}{2} \arcsin \left( \frac{x}{R} \right) + \frac{1}{2} \times \frac{2x \sqrt{R^2 - x^2}}{R^2} + C \] Simplify: \[ I = \frac{R^2}{2} \arcsin \left( \frac{x}{R} \right) + \frac{x \sqrt{R^2 - x^2}}{R^2} + C \] **Multiplying numerator and denominator of the second term by \(R^2\):** \[ I = \frac{R^2}{2} \arcsin \left( \frac{x}{R} \right) + \frac{x \sqrt{R^2 - x^2}}{R^2} + C \] Alternatively, write as: \[ \boxed{ \int \sqrt{R^2 - x^2} \, dx = \frac{R^2}{2} \arcsin \left(\frac{x}{R}\right) + \frac{x}{2} \sqrt{R^2 - x^2} + C } \] *(Note: Both forms are equivalent; the second is often more convenient for evaluation.)* --- ### **Part 3: Domain of \(f(x)\)** Since: \[ f(x) = \sqrt{R^2 - x^2} \] The radicand must be non-negative: \[ R^2 - x^2 \ge 0 \implies -R \le x \le R \] **Answer:** \[ \boxed{ \text{Domain: } [-R, R] } \] --- ### **Part 4: Graph and shaded area** - The graph of \(f(x) = \sqrt{R^2 - x^2}\) is the **upper semicircle** of radius \(R\), centered at the origin. - **Sketch instructions:** - Draw the circle \(x^2 + y^2 = R^2\). - Shade the **upper half**: \(y = \sqrt{R^2 - x^2}\), for \(x \in [-R, R]\). - The shaded region under the curve from \(-R\) to \(R\) represents **half the area of the circle**. --- ### **Part 5: Calculate the definite integral using the Fundamental Theorem** \[ \int_{-R}^{R} \sqrt{R^2 - x^2} \, dx \] - Using the indefinite integral \(F(x)\) from Part 2: \[ F(x) = \frac{R^2}{2} \arcsin \left(\frac{x}{R}\right) + \frac{x}{2} \sqrt{R^2 - x^2} \] - **Evaluate at the bounds:** At \(x = R\): \[ \arcsin(1) = \frac{\pi}{2} \] \[ \sqrt{R^2 - R^2} = 0 \] \[ F(R) = \frac{R^2}{2} \times \frac{\pi}{2} + \frac{R}{2} \times 0 = \frac{\pi R^2}{4} \] At \(x = -R\): \[ \arcsin(-1) = -\frac{\pi}{2} \] \[ \sqrt{R^2 - R^2} = 0 \] \[ F(-R) = \frac{R^2}{2} \times \left(- \frac{\pi}{2}\right) + \frac{-R}{2} \times 0 = - \frac{\pi R^2}{4} \] - **Compute the definite integral:** \[ \int_{-R}^{R} \sqrt{R^2 - x^2} \, dx = F(R) - F(-R) = \frac{\pi R^2}{4} - \left(- \frac{\pi R^2}{4}\right) = \frac{\pi R^2}{2} \] **Final answer:** \[ \boxed{ \int_{-R}^{R} \sqrt{R^2 - x^2} \, dx = \frac{\pi R^2}{2} } \] --- ### **Part 6: Geometric significance** - The result \(\frac{\pi R^2}{2}\) represents **half of the area of a circle** with radius \(R\): \[ \text{Area of full circle} = \pi R^2 \] - Since the integral computes the area under the upper semicircle: \[ \boxed{ \text{Area of upper semicircle} = \frac{\pi R^2}{2} } \] - **Multiplying** this area by 4 (as hinted), we get the **area of the entire circle**: \[ 4 \times \frac{\pi R^2}{2} = 2 \pi R^2 \] But note, this is double the actual circle area unless the problem refers to something else. Likely, the key point is: > The area under the upper half of the circle (semicircle) corresponds to \(\frac{\pi R^2}{2}\). --- ## **Summary of final answers:** 1. **Derivative:** \[ f'(x) = \frac{-x}{\sqrt{R^2 - x^2}} \] 2. **Indefinite integral:** \[ \int \sqrt{R^2 - x^2} \, dx = \frac{R^2}{2} \arcsin \left(\frac{x}{R}\right) + \frac{x}{2} \sqrt{R^2 - x^2} + C \] 3. **Domain:** \[ x \in [-R, R] \] 4. **Graph:** - Semicircle \(x^2 + y^2 = R^2\), upper half. - Shaded area under \(f(x)\), from \(-R\) to \(R\). 5. **Definite integral:** \[ \boxed{ \int_{-R}^{R} \sqrt{R^2 - x^2} \, dx = \frac{\pi R^2}{2} } \] 6. **Geometric significance:** - The integral equals **half the area of the circle** of radius \(R\). --- Please let me know if you'd like clarification on any step!