Using integrals:

Since the |x-5| is equal to a positive value when x ranges from 5 to 10 but equal to a negative value from 0 to 5, we have to split the integral into two parts. Notice in the previous answer how the graph creates two triangles.

Here’s what we have: ##int_0^5- (x-5) dx## + ##int_5^10 (x-5) dx##

When you integrate using your you obtain: ##-1/2 x^2 + 5x## from 0 to 5 + ##1/2 x^2 – 5x## from 5 to 10.

Substituting upper bound minus lower bound for the first integral results in ##[-1/2(25)+5(5)] – [-1/2 (0) +5(0)]## which equals ##25/2##.

Doing the same for the second integral gives us ##[1/2(100)-5(10)] – [1/2(25)-5(5)]## which equals ##25/2##.

Add the two integral values for the total area: ##25/2+25/2=25##

Hope this helps.