Question

Evaluate the integral \(\int_1^{\sqrt{3}} \frac{3}{1+x^2}\).

(a) \(\frac{π}{2}\)

(b) \(\frac{π}{4}\)

(c) \(\frac{π}{3}\)

(d) \(\frac{π}{6}\)

This question was addressed to me during an internship interview.

The origin of the question is Fundamental Theorem of Calculus-2 in portion Integrals of Mathematics – Class 12

Answer 1

Correct choice is (b) \(\frac{π}{4}\)

The best I can explain: Let \(I=\int_1^{√3} \frac{3}{1+x^2}\)

F(x)=\(\int \frac{3}{1+x^2}dx\)

=3\(\int \frac{1}{1+x^2} \,dx\)

=3 tan^-1⁡x

Applying the limits, we get

I=F(\(\sqrt{3}\))-F(1)

=3 tan^-1⁡\(\sqrt{3}\)-3 tan^-1⁡1

\(3(\frac{π}{3})-\frac{3π}{4}=\frac{4π-3π}{4}=\frac{π}{4}\).