Question

By using the properties of definite integrals,evaluate the integrals: $\int^8_2 {\arrowvert x-5\arrowvert} \, dx$

Answer 1

Answer:

The value of the given integral is 14

Step-by-step explanation:

Concept used:

$\int\limits^b_a{f(x)}\:dx=\int\limits^c_a{f(x)}\:dx+\int\limits^b_c{f(x)}\:dx$

Now,

$\int\limits^8_2{|x-2|}\:dx\\\\=\int\limits^5_2{|x-2|}\:dx+\int\limits^8_5{|x-2|}\:dx$

$=\int\limits^5_2{-(x-2)}\:dx+\int\limits^8_5{x-2}\:dx\\\\=\int\limits^5_2{-x+2}\:dx+\int\limits^8_5{x-2}\:dx$

$=[\frac{-x^2}{2}+2x]^5_2+[\frac{x^2}{2}-2x]^8_5\\\\=(\frac{-5^2}{2}+2(5))-(\frac{-2^2}{2}+2(2))+(\frac{8^2}{2}-2(8))-(\frac{5^2}{2}-2(5))$

$=(\frac{-25}{2}+10)-(\frac{-4}{2}+4)+(\frac{64}{2}-16)-(\frac{25}{2}-10)\\\\=(\frac{-25+20}{2})-(\frac{-4+8}{2})+(\frac{64-32}{2})-(\frac{25-20}{2})$

$=(\frac{-5+}{2})-(\frac{4}{2})+(\frac{32}{2})-(\frac{5}{2})\\\\=\frac{-4+32}{2}\\\\=\frac{28}{2}\\\\=14$