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  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. Integral 0 to 4 |x - 1 |dx

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

    Q18.    \int_0^4 |x-1|dx

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We have,                          I\ =\ \int_{0}^4|x-1|dx

For opening the modulus we need to define the bracket :

           If     (x - 1)  < 0    then  x belongs to (0, 1).  And if (x - 1) > 0 then x belongs to (1, 4).

So the integral becomes:-              

                                         I\ =\ \int_{0}^{1} -(x-1)dx\ +\ \int_{1}^{4} (x-1)dx

or                                     I\ =\ \left [ x\ -\ \frac{x^2}{2}\ \right ]^{1} _{0}\ +\ \left [ \frac{x^2}{2}\ -\ x \right ]^{4} _{1}

This gives                       I\ =\ 5 

Posted by

Devendra Khairwa

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