Question

The area bounded by the curve \( y = \sin x \) and the x-axis between \( x = 0 \) and \( x = \frac{3\pi}{2} \) is ............. sq. units. (answer in integer)

Hint:

The area under a sine curve can be calculated by evaluating the definite integral of \( \sin x \) over the given limits.

Answer 1

Step 1: Understanding the problem.
We are asked to find the area bounded by the curve \( y = \sin x \) and the x-axis between \( x = 0 \) and \( x = \frac{3\pi}{2} \). The area under the curve from \( x = 0 \) to \( x = \frac{3\pi}{2} \) can be found using the definite integral: \[ \text{Area} = \int_0^{\frac{3\pi}{2}} \sin x \, dx \] Step 2: Solving the integral.
We need to evaluate the integral: \[ \int \sin x \, dx = -\cos x \] Now, applying the limits: \[ \text{Area} = - \cos \left( \frac{3\pi}{2} \right) + \cos (0) \] \[ \cos \left( \frac{3\pi}{2} \right) = 0 \quad \text{and} \quad \cos (0) = 1 \] So, the area is: \[ \text{Area} = - (0) + 1 = 1 \] Final Answer: \[ \boxed{1} \]