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

Concept: Recall the standard integral: \[ \int \frac{1}{1+x^2}\,dx = \tan^{-1}x+C \] For definite integrals: \[ \int_a^b f(x)\,dx = F(b)-F(a) \] where \(F(x)\) is antiderivative of \(f(x)\).

Step 1: Integrating the function. Given: \[ I = \int_0^1 \frac{1}{1+x^2}\,dx \] Using standard result: \[ \int \frac{1}{1+x^2}\,dx = \tan^{-1}x \] Thus: \[ I = \left[\tan^{-1}x\right]_0^1 \]

Step 2: Applying limits. Substitute upper and lower limits: \[ I = \tan^{-1}(1)-\tan^{-1}(0) \] We know: \[ \tan^{-1}(1)=\frac{\pi}{4} \] and \[ \tan^{-1}(0)=0 \] Hence: \[ I = \frac{\pi}{4} \] Final Answer: \[ \boxed{(B)\ \frac{\pi}{4}} \]