Get Answers to all your Questions

header-bg qa

Choose the correct answer in Exercises 20 and 21.

    Q20.    The value of is \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx is 

                    (A)    0

                    (B)    2

                    (C)    \pi

                    (D)    1

Answers (1)

best_answer

We have                             

                                                I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx

This can be written as :

                                               I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}x^3dx +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} x\cos x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} \tan^5 x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} 1dx

Also if a function is even function then       \int_{-a}^{a}f(x)\ dx\ =\ 2\int_{0}^{a}f(x)\ dx

And if the function is an odd function then :                \int_{-a}^{a}f(x)\ dx\ =\ 0

Using the above property I become:- 

                                                 I\ =\ 0+0+0+ 2\int_{0}^{\frac{\Pi }{2}}1.dx

or                                              I\ =\ 2\left [ x \right ]^{\frac{\Pi }{2}}_0

or                                              I\ =\ \Pi

Posted by

Devendra Khairwa

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads