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  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. Integral - 5 to 5 |x + 2| dx

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

    Q5.    \int_{-5}^5|x+2|dx

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We have,                          I\ =\ \int_{-5}^5|x+2|dx

For opening the modulus we need to define the bracket :

           If     (x + 2)  < 0    then  x belongs to (-5, -2).  And if (x + 2) > 0 then x belongs to (-2, 5).

So the integral becomes :-              

                                         I\ =\ \int_{-5}^{-2} -(x+2)dx\ +\ \int_{-2}^{5} (x+2)dx

or                                     I\ =\ -\left [ \frac{x^2}{2}\ +\ 2x \right ]^{-2} _{-5}\ +\ \left [ \frac{x^2}{2}\ +\ 2x \right ]^{5} _{-2}

This gives                       I\ =\ 29 

Posted by

Devendra Khairwa

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