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  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. Integral 2 to 8 Absolute value of x minus 5 dx

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

    Q6.    \int_2^8|x-5|dx

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We have,                          I\ =\ \int_{2}^8|x-5|dx

For opening the modulus we need to define the bracket :

           If     (x - 5)  < 0    then  x belongs to (2, 5).  And if (x - 5) > 0 then x belongs to (5, 8).

So the integral becomes:-              

                                         I\ =\ \int_{2}^{5} -(x-5)dx\ +\ \int_{5}^{8} (x-5)dx

or                                     I\ =\ -\left [ \frac{x^2}{2}\ -\ 5x \right ]^{5} _{2}\ +\ \left [ \frac{x^2}{2}\ -\ 5x \right ]^{8} _{5}

This gives                       I\ =\ 9 

Posted by

Devendra Khairwa

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