definite integral of the function

[ssmmss]

Is the definite integral of the function, attached JPEG, from 2 to 8 --> 20, simply by counting the number of blocks.

 

[pka]

Is the definite integral of the function, attached JPEG, from 2 to 8 --> 20, simply by counting the number of blocks.

Add the areas of two rectangles: \(\displaystyle 2\times 2~\&~4\times 4\).

 

[HallsofIvy]

Oops- I was thinking this was from 0 to 8!

You can also do this as the area of two rectangles as pka suggests or, since this was posted under "Calculus", you can think of the upper boundary as the piecewise function f(x)= 2 if x< 4, f(x)= 4 if x> 4. Then the area is \(\displaystyle \int_2^8 f(x)dx= \int_2^4 2dx+ \int_4^8 4dx= \left[2x\right]_2^4+ \left[4t\right]_4^8= 2(4- 2)+ 4(8- 4)= 4+ 16= 20\).

 

[lookagain]

... the piecewise function f(x)= 2 if x< 4, f(x)= 4 if x> 4.

Then the area is \(\displaystyle \int_2^8 f(x)dx = \)

\(\displaystyle \int_2^4 2dx \ + \ \int_4^8 4dx = \)

\(\displaystyle \left[2x\right]_2^4 \ + \ \left[4t\right]_4^8 = \ \ \ \ \ \) <------ You meant\(\displaystyle \ \ \ \left[2x\right]_2^4 \ + \ \left[4x\right]_4^8 \)


\(\displaystyle 2(4 - 2) \ + \ 4(8 - 4) = \)

\(\displaystyle 4 + 16 = \)

\(\displaystyle 20\).

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