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Choose the correct answer in Exercises 20 and 21.

    Q21.    \int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}

                (A)    \frac{\pi}{3}

                (B)    \frac{2\pi}{3}

                (C)    \frac{\pi}{6}

                (D)    \frac{\pi}{12}

Answers (1)

best_answer

Given definite integral \int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}

Consider \int \frac{dx}{1 +x^2} = \tan^{-1}x

we have then the function of x, as f(x) = \tan^{-1}x

By applying the second fundamental theorem of calculus, we will get

\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2} = f(\sqrt3) - f(1)

= \tan^{-1}\sqrt{3} - \tan^{-1}1

=\frac{\pi}{3} - \frac{\pi}{4}

= \frac{\pi}{12}

Therefore the correct answer is D.

Posted by

Divya Prakash Singh

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