Evaluate the integral ∫(0 to π) sin(x) dx. (2021)

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Question: Evaluate the integral ∫(0 to π) sin(x) dx. (2021)

Options:

Correct Answer: 2

Exam Year: 2021

Solution:

∫(0 to π) sin(x) dx = [-cos(x)] from 0 to π = -(-1 - 1) = 2.

Evaluate the integral ∫(0 to π) sin(x) dx. (2021)

Evaluate the integral ∫(0 to π) sin(x) dx. (2021)

Step 1: Identify the integral you need to evaluate: ∫(0 to π) sin(x) dx.
Step 2: Find the antiderivative of sin(x). The antiderivative is -cos(x).
Step 3: Write down the limits of integration, which are from 0 to π.
Step 4: Apply the Fundamental Theorem of Calculus: Evaluate -cos(x) at the upper limit (π) and the lower limit (0).
Step 5: Calculate -cos(π): This equals -(-1) = 1.
Step 6: Calculate -cos(0): This equals -1.
Step 7: Subtract the value at the lower limit from the value at the upper limit: 1 - (-1) = 1 + 1 = 2.
Step 8: Conclude that the value of the integral ∫(0 to π) sin(x) dx is 2.

Definite Integral – The question tests the understanding of evaluating definite integrals, specifically using the Fundamental Theorem of Calculus.
Trigonometric Functions – The integral involves the sine function, requiring knowledge of its properties and antiderivative.
Limits of Integration – Understanding how to apply the limits of integration correctly when evaluating the definite integral.