Question

The area under the curve $f(x) = \sin x$ in $[0, 2\pi]$ is

• A. $1$
• B. $3$
• C. $-4$
• D. $4$

Hint:

Be careful with the wording! If the question asks for the definite integral $\int_{0}^{2\pi} \sin x dx$, the answer is $0$ because the positive and negative parts cancel out. "Area" implies the physical space, which is always positive.

Answer 1

The Correct Option is D

1. Analyzing the Function Behavior: The function $f(x) = \sin x$ behaves as follows in $[0, 2\pi]$:

• $[0, \pi]$: $\sin x \geq 0$ (Positive region)

• $[\pi, 2\pi]$: $\sin x \leq 0$ (Negative region)

2. Setting up the Total Area Integral: $$\text{Total Area} = \int_{0}^{\pi} \sin x dx + \left| \int_{\pi}^{2\pi} \sin x dx \right|$$

3. Calculating the Parts: For the first half: $$\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$$ For the second half: $$\int_{\pi}^{2\pi} \sin x dx = [-\cos x]_{\pi}^{2\pi} = -(\cos 2\pi - \cos \pi) = -(1 - (-1)) = -2\lt strong\gt 4. Summing the Absolute Magnitudes:\lt /strong\gt \text{Total Area} = 2 + |-2| = 2 + 2 = 4$$ Thus, the total geometric area is 4 square units.