3.7 Q-12

Question Statement

Find the area between the x-axis and the curve $y = \sin 2x$ from $x = 0$ to $x = \frac{\pi}{3}$.

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Background and Explanation

To solve this problem, we need to:

• Integrate the function $y = \sin 2x$ over the given interval from $x = 0$ to $x = \frac{\pi}{3}$.
• Recognize that the integral of sine functions is related to cosine functions, and we will use this relationship to evaluate the area.

We will use the definite integral of $\sin 2x$ to find the area under the curve, as the graph of $\sin 2x$ is above the x-axis in this interval.

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Solution

1. Set up the integral for the area: The area between the curve and the x-axis is given by the integral:

$$\text{Area} = \int_{0}^{\frac{\pi}{3}} \sin 2x , dx$$

2. Apply the integral of sine: The integral of $\sin 2x$ requires a constant factor for the derivative of $2x$. We can rewrite the integral as:

$$\text{Area} = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} -2 \sin 2x , dx$$

This step ensures we account for the factor of 2 inside the sine function.

3. Integrate the function: The integral of $\sin 2x$ is $-\frac{1}{2} \cos 2x$, so:

$$\text{Area} = -\frac{1}{2} \cos 2x \bigg|_0^{\frac{\pi}{3}}$$

4. Evaluate the definite integral: Now substitute the upper and lower limits into the cosine function:

$$\text{Area} = -\frac{1}{2} \left[ \cos \left( 2 \times \frac{\pi}{3} \right) - \cos (0) \right]$$

We know that $\cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2}$ and $\cos (0) = 1$, so:

$$\text{Area} = -\frac{1}{2} \left[ -\frac{1}{2} - 1 \right]$$

Simplifying:

$$\text{Area} = -\frac{1}{2} \left[ \frac{-1 - 2}{2} \right] = \frac{3}{4}$$

Thus, the area under the curve is:

$$\text{Area} = \frac{3}{4} , \text{square units}$$

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Key Formulas or Methods Used

• Integral of sine function:

$$\int \sin 2x , dx = -\frac{1}{2} \cos 2x$$

• Definite integral:

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

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Summary of Steps

1. Set up the integral: $\int_{0}^{\frac{\pi}{3}} \sin 2x , dx$.
2. Use the factor $\frac{1}{2}$ to account for the derivative of $2x$.
3. Integrate: $\text{Area} = -\frac{1}{2} \cos 2x$.
4. Substitute the limits of integration: $\text{Area} = \frac{3}{4}$.