3.7 Q-4

Question Statement

Find the area bounded by the cosine function from $x = -\frac{\pi}{2}$ to $x = \frac{\pi}{2}$.

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Background and Explanation

This problem involves finding the area under the curve $y = \cos(x)$ between the limits $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$. The area under a curve is calculated using definite integration. The cosine function oscillates between -1 and 1, so we are interested in the integral of this function over the given interval.

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Solution

We need to compute the following integral to find the area under the curve:

$$A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(x) , dx$$

Step 1: Integrate the cosine function

The integral of $\cos(x)$ is $\sin(x)$. Applying this:

$$\int \cos(x) , dx = \sin(x)$$

Thus, the area becomes:

$$A = \sin(x) \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$$

Step 2: Evaluate the integral at the limits

Now, we evaluate $\sin(x)$ at the limits $x = \frac{\pi}{2}$ and $x = -\frac{\pi}{2}$:

$$A = \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right)$$

Since $\sin\left(\frac{\pi}{2}\right) = 1$ and $\sin\left(-\frac{\pi}{2}\right) = -1$, we get:

$$A = 1 - (-1) = 2$$

Thus, the area under the curve is 2 square units.

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Key Formulas or Methods Used

• Definite Integral: The area under the curve from $x = a$ to $x = b$ is given by:

$$A = \int_{a}^{b} f(x) , dx$$

• Integral of Cosine: The integral of $\cos(x)$ is:

$$\int \cos(x) , dx = \sin(x)$$

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Summary of Steps

1. Set up the integral: $A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(x) , dx$

2. Integrate the cosine function: $\int \cos(x) , dx = \sin(x)$

3. Evaluate the integral at the limits: $A = \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right)$

4. Simplify: $A = 1 - (-1) = 2$

5. The area is 2 square units.