Intergation

[bamby]

Can you help me with this problem (attached):

 

[BigGlenntheHeavy]

$\displaystyle Find \ \int_{0}^{6}|x-4|dx$

$\displaystyle |x-4| \ = \ x-4 \ if \ x-4 \ is \ greater \ or \ equal \ to \ zero, \ x \ greater \ or \ equal \ 4.$

$\displaystyle |x-4| \ = \ -(x-4) \ if \ x-4 \ is \ less \ than \ zero, \ x \ is \ less \ than \ 4.$

$\displaystyle Ergo, \ \int_{0}^{6}|x-4|dx \ = \ \int_{4}^{6}(x-4)dx \ + \ \int_{0}^{4}(4-x)dx \ = \ 2+8 \ = \ 10. \ QED$

Note: for you aspiring engineers out there, the absolute value is the way to go if you were going to pave this total area in cement one foot deep, as knowing the area is 10, you would need 10 cubic feet of cement.

 

[bamby]

Thanks!

 

[soroban]

Hello, bamby!

If you sketch the graph, the answer is obvious . . .

$\displaystyle \int^6_0|x-4|\,dx$

$\displaystyle \text{The graph of }\:y \:=\:|x|\:\text{ is a "V" with its vertex at the origin.}$

$\displaystyle \text{The graph of }\:y \:=\:|x-4|\:\text{ is the "V" moved 4 units to the right.}$

$\displaystyle \text{The integral asks for the area under the graph from }x=0\text{ to }x = 6.$

        |
   (0,4)*
        |:*
        |:::*
        |:::::*           *(6,2)
        |:::::::*       *:|
        |:::::::::*   *:::|
    - - + - - - - - * - - + - -
        |           4     6

Got it?

 

[bamby]

Yes, thank you!