3.7 Q-11

Question Statement

Find the area between the x-axis and the curve $y = \cos \frac{x}{2}$ from $x = -\pi$ to $x = \pi$.

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Background and Explanation

This problem involves finding the area under a trigonometric curve. To do this, we will:

• Integrate the function $y = \cos \frac{x}{2}$ from $x = -\pi$ to $x = \pi$.
• Recognize that the graph of $\cos \frac{x}{2}$ remains above the x-axis over this interval, simplifying the process of calculating the area.

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Solution

1. Set up the integral: We need to calculate the area under the curve $y = \cos \frac{x}{2}$ from $x = -\pi$ to $x = \pi$. The area is given by the integral:

$$\text{Area} = \int_{-\pi}^{\pi} \cos \frac{x}{2} , dx$$

2. Use substitution for easier integration: To solve the integral, we can simplify by applying substitution. Let:

$$u = \frac{x}{2}, \quad \text{so} \quad du = \frac{1}{2} , dx$$

This leads to:

$$\text{Area} = 2 \int_{-\pi/2}^{\pi/2} \cos u , du$$

(since the limits of integration change accordingly).

3. Integrate the function: The integral of $\cos u$ is $\sin u$. So, we integrate:

$$\text{Area} = 2 \left[ \sin u \right]_{-\pi/2}^{\pi/2}$$

4. Evaluate the definite integral: Now, substitute the limits:

$$\text{Area} = 2 \left[ \sin \frac{\pi}{2} - \sin \left( -\frac{\pi}{2} \right) \right]$$

We know that $\sin \frac{\pi}{2} = 1$ and $\sin \left( -\frac{\pi}{2} \right) = -1$, so:

$$\text{Area} = 2 \left[ 1 - (-1) \right] = 2 \times 2 = 4$$

Thus, the area under the curve is:

$$\text{Area} = 4 , \text{square units}$$

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Key Formulas or Methods Used

• Substitution for integration:

$$u = \frac{x}{2}, \quad du = \frac{1}{2} dx$$

• Standard integral of cosine:

$$\int \cos u , du = \sin u$$

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Summary of Steps

1. Set up the integral for the area: $\int_{-\pi}^{\pi} \cos \frac{x}{2} , dx$.
2. Apply substitution: $u = \frac{x}{2}$, so $du = \frac{1}{2} dx$.
3. Simplify and solve the integral: $2 \int_{-\pi/2}^{\pi/2} \cos u , du$.
4. Evaluate the definite integral: $\text{Area} = 4$.