Question

The area bounded by the curve \[ y = \sin\left(\frac{x}{3}\right), \quad x \text{ axis}, \quad \text{the lines } x = 0 \text{ and } x = 3\pi \] is:

• A. 1 sq. units
• B. 6 sq. units
• C. 3 sq. units
• D. 9 sq. units

Hint:

When calculating areas under trigonometric curves, the use of substitution and trigonometric identities simplifies the integration process.

Answer 1

The Correct Option is B

We need to calculate the area under the curve: \[ \int_0^{3\pi} \sin\left(\frac{x}{3}\right) \, dx \] For the given function: \[ f(x) = \sin\left(\frac{x}{3}\right) \] The integral becomes: \[ \int_0^{3\pi} \sin\left(\frac{x}{3}\right) \, dx \] Using the substitution \( u = \frac{x}{3} \), so \( du = \frac{dx}{3} \) or \( dx = 3du \). When \( x = 0, u = 0 \) and when \( x = 3\pi, u = \pi \). The integral becomes: \[ 3 \int_0^\pi \sin(u) \, du = 3 \left[ -\cos(u) \right]_0^\pi \] \[ = 3 \left[ -\cos(\pi) + \cos(0) \right] = 3 \left[ 1 + 1 \right] = 6 \] Thus, the area is 6 square units.