Question

Evaluate: $\int_0^{\pi/2}\sin x \, dx$

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( \frac{\pi}{2} \)

Hint:

The area of a single standard loop quadrant of either a sine or cosine curve between consecutive axis intercepts (e.g., from \(0\) to \(\frac{\pi}{2}\)) is always exactly \(1\text{ unit}^2\). Remembering this geometric curve property makes evaluating simple definite trigonometric integrations instantaneous.

Answer 1

The Correct Option is B

Concept: According to the Fundamental Theorem of Calculus, the value of a definite integral is computed by evaluating the anti-derivative of the function at the upper limit and subtracting its value at the lower limit: \[ \int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a) \] Where \(F'(x) = f(x)\). The basic standard formula for the anti-derivative of \(\sin x\) is: \[ \int \sin x \, dx = -\cos x \]

Step 1: Finding the definite limits execution.
Apply the anti-derivative tracking boundaries from \(0\) to \(\frac{\pi}{2}\): \[ \int_0^{\pi/2} \sin x \, dx = \Big[-\cos x\Big]_0^{\pi/2} \]

Step 2: Substituting the boundary values.
Evaluate at the upper limit and lower limit: \[ \int_0^{\pi/2} \sin x \, dx = -\left( \cos\frac{\pi}{2} - \cos 0 \right) \] We know the standard exact exact trigonometric values are \(\cos\frac{\pi}{2} = 0\) and \(\cos 0 = 1\): \[ \int_0^{\pi/2} \sin x \, dx = -(0 - 1) = 1 \]