Question

Evaluate: \[ \int_{0}^{1} \frac{1}{1+x^2}\,dx \]

• A. \(\frac{\pi}{2}\)
• B. \(\frac{\pi}{4}\)
• C. \(1\)
• D. \(\ln 2\)

Hint:

Remember: \[ \int \frac{1}{1+x^2}\,dx = \tan^{-1}x+C \] and: \[ \tan^{-1}(1)=\frac{\pi}{4} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires evaluating a definite integral.
A definite integral represents the signed area under a curve between two points \( a \) and \( b \).
The function being integrated is \( f(x) = \frac{1}{1 + x^2} \).
According to the Fundamental Theorem of Calculus, if \( F(x) \) is the antiderivative of \( f(x) \), then:
\[ \int_{a}^{b} f(x) dx = F(b) - F(a) \]
Step 2: Key Formula or Approach:
The standard antiderivative of \( \frac{1}{1 + x^2} \) is \( \tan^{-1} x \).
Limits of integration are \( a = 0 \) and \( b = 1 \).
Step 3: Detailed Explanation:
Step 1: Find the indefinite integral (antiderivative).
We know from standard calculus tables that:
\[ \int \frac{1}{1 + x^2} dx = \tan^{-1} x + C \]
Step 2: Apply the limits for the definite integral.
The expression to evaluate is:
\[ \left[ \tan^{-1} x \right]_{0}^{1} \]
Step 3: Calculate the value at the upper limit minus the value at the lower limit.
\[ \text{Result} = \tan^{-1}(1) - \tan^{-1}(0) \]
Step 4: Recall trigonometric values for inverse functions.
We ask: For what angle \( \theta \) is \( \tan \theta = 1 \) in the principal range?
The answer is \( \theta = \frac{\pi}{4} \).
We ask: For what angle \( \theta \) is \( \tan \theta = 0 \)?
The answer is \( \theta = 0 \).
Step 5: Substitute these values into the expression.
\[ \text{Result} = \frac{\pi}{4} - 0 \]
\[ \text{Result} = \frac{\pi}{4} \]
Step 4: Final Answer:
The value of the definite integral is \( \frac{\pi}{4} \).
This matches Option (B).