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- WE WANT TO EVALUATE THE GIVEN DEFINITE INTEGRAL
USING A GEOMETRIC FORMULA.
SO IF A GIVEN FUNCTION F OF X IS CONTINUOUS AND NON-NEGATIVE
ON A CLOSED INTERVAL FROM A TO B
THEN THE AREA UNDER THAT FUNCTION AND ABOVE THE X-AXIS
IS EQUAL TO THE DEF INTEGRAL OF F OF X FROM A TO B.
SO LOOKING AT OUR DEFINITE INTEGRAL THE FIRST THING
WE SHOULD RECOGNIZE IS THAT THE CLOSED INTERVAL
WOULD BE FROM 0 TO 6 GIVEN BY THE LIMITS OF INTEGRATION
AND THAT OUR FUNCTION F OF X IS EQUAL TO THE ABSOLUTE VALUE
OF THE QUANTITY X - 4.
LET'S GO AHEAD AND JUST CALL OUR FUNCTION
Y = THE ABSOLUTE VALUE OF X - 4.
SO THE NEXT THING I'VE NOTICED IS THAT THE FUNCTION VALUE
WILL ALWAYS BE NON-NEGATIVE,
AND THEREFORE THIS MEANS IF WE GRAPH THIS FUNCTION
ON THIS CLOSED INTERVAL
AND THEN FIND THE AREA BELOW THE FUNCTION
AND ABOVE THE X-AXIS IT WILL EQUAL THIS DEFINITE INTEGRAL.
SO LET'S FOCUS ON GRAPHING THIS ABSOLUTE VALUE FUNCTION.
WE SHOULD RECOGNIZE THIS WILL MAKE A V-SHAPED GRAPH,
AND THE V WILL OPEN UP.
LET'S START BY DETERMINING THE FUNCTION VALUES
AT X = 0 AND X = 6.
SO IF X = 0 NOTICE HOW WE WOULD HAVE Y = THE ABSOLUTE VALUE
OF 0 - 4.
THE ABSOLUTE VALUE OF -4 IS 4,
SO OUR FUNCTION CONTAINS THE POINT (0,4) HERE.
NEXT, WHEN X = 6 WE WOULD HAVE Y = THE ABSOLUTE VALUE OF 6 - 4
AND THE ABSOLUTE VALUE OF 2 IS 2.
SO OUR FUNCTION ALSO CONTAINS THE POINT (6,2) WHICH IS HERE.
BUT PROBABLY THE MOST IMPORTANT POINT WE NEED TO FIND
IS WHERE THE GRAPH TURNS OR CHANGES DIRECTION,
AND THIS WILL OCCUR WHERE THE QUANTITY X - 4
IS EQUAL TO 0 OR AT THE X-INTERCEPT.
NOTICE IF WE SET X EQUAL TO 4
WE WOULD HAVE Y = THE ABSOLUTE VALUE OF 4 - 4
WHICH IS THE ABSOLUTE VALUE OF 0 OR 0.
SO THIS GRAPH IS GOING TO CHANGE DIRECTIONS AT THIS POINT,
OR THIS WILL BE THE TIP OF THE V.
SO THE POINT (4,0) WOULD BE HERE ON THE X-AXIS,
AND THEREFORE THE GRAPH OF THE ABSOLUTE VALUE FUNCTION
MOVES UP IN THIS DIRECTION AND THEN MOVES UP IN THIS DIRECTION.
BUT REMEMBER OUR GOAL IS ONLY TO FIND THE AREA UNDER THE FUNCTION
AND ABOVE THE X-AXIS ON THE INTERVAL FROM 0 TO 6.
SO WE'RE GOING TO START HERE AND STOP HERE.
SO IF WE SHADE THE AREA UNDER THE FUNCTION ON THIS INTERVAL
IT WOULD LOOK LIKE THIS.
BECAUSE THESE TWO AREAS ARE TRIANGLES,
IF WE FIND THE SUM OF THESE TWO AREAS
IT WILL BE EQUAL TO THE GIVEN DEFINITE INTEGRAL.
LET'S GO AHEAD AND CALL THIS AREA SUB 1
AND CALL THIS AREA SUB 2.
AND BECAUSE THESE ARE TRIANGLES
WE CAN USE THE AREA FORMULA AREA = 1/2 BASE x HEIGHT,
OR IF WE WANT, BASE x HEIGHT DIVIDED BY 2.
NOTICE HOW THIS FIRST TRIANGLE HAS A BASE OF 4
AND ALSO A HEIGHT OF 4,
AND THE SECOND TRIANGLE HAS A BASE OF 2 AND A HEIGHT OF 2.
SO NOW WE KNOW THAT THE INTEGRAL OF THE ABSOLUTE VALUE OF X - 4
ON THE INTERVAL FROM 0 TO 6
WILL BE EQUAL TO AREA SUB 1 + AREA SUB 2.
SO AREA SUB 1 IS GOING TO BE EQUAL TO BASE x HEIGHT
DIVIDED BY 2 OR 4 x 4 DIVIDED BY 2.
THAT'LL BE 16 DIVIDED 2 OR 8
AND AREA SUB 2 IS EQUAL TO 2 x 2 DIVIDED BY 2.
4 DIVIDED BY 2 IS 2.
SO THE DEFINITE INTEGRAL IS EQUAL TO 8 + 2
WHICH IS EQUAL TO 10.
I HOPE YOU FOUND THIS EXPLANATION HELPFUL.