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- WE WANT TO EVALUATE THE GIVEN DEFINITE INTEGRAL
USING A GEOMETRIC FORMULA.
SO TO REVIEW, IF A FUNCTION F OF X IS CONTINUOUS
AND NON-NEGATIVE ON A CLOSED INTERVAL FROM A TO B,
THEN THE AREA UNDER THE FUNCTION AND ABOVE THE X-AXIS,
AS WE SEE SHADED HERE, IS EQUAL TO THE DEFINITE INTEGRAL F OF X
FROM A TO B.
SO GOING BACK TO OUR EXAMPLE,
THE FIRST THING WE SHOULD RECOGNIZE IS THAT
THE CLOSED INTERVAL WILL BE FROM 0 TO 6
GIVEN BY THE LIMITS OF INTEGRATION,
AND OUR FUNCTION F OF X WOULD BE EQUAL TO 3/2X - 3.
LET'S GO AHEAD AND CALL OUR FUNCTION Y = 3/2X - 3.
SO LET'S GO AHEAD AND GRAPH OUR LINEAR FUNCTION
ON THE CLOSED INTERVAL AND SEE IF IT'S NON-NEGATIVE.
WE SHOULD RECOGNIZE THIS IS IN SLOPED INTERCEPT
OR THE FORM Y = MX + B, SO THE Y-INTERCEPT IS -3 WHICH IS HERE.
SO WE CAN ALREADY SEE THE FUNCTION
IS NOT GOING TO BE NON-NEGATIVE,
BUT FROM HERE, BECAUSE THE SLOPE IS 3/2
WE'LL GO UP 3 UNITS AND RIGHT 2 UNITS.
LET'S DO THAT AGAIN, UP 3, RIGHT 2.
LET'S DO THAT AGAIN, UP 3, RIGHT 2.
SO THIS IS THE GRAPH OF OUR LINEAR FUNCTION,
AND WE'RE ONLY CONCERNED ABOUT THIS FUNCTION
ON THE INTERVAL FROM 0 TO 6,
MEANING STARTING HERE AND ENDING HERE.
SO EVEN THOUGH A PIECE OF THIS FUNCTION IS NEGATIVE,
WE CAN STILL ACTUALLY USE GEOMETRIC FORMULAS
TO EVALUATE THIS DEFINITE INTEGRAL.
THE AREA BOUNDED BY THE FUNCTION IN THE X-AXIS
WOULD BE HERE AND HERE.
AND BECAUSE THE FUNCTION IS NEGATIVE ON THE INTERVAL
FROM 0 TO 2 WE CAN STILL FIND THE AREA OF THIS TRIANGLE
USING THE FORMULA AREA = 1/2 BASE x HEIGHT OR BASE x HEIGHT
DIVIDED BY 2 AND THEN JUST MAKE THAT NEGATIVE
BECAUSE THE FUNCTION IS NEGATIVE ON THE INTERVAL.
AND THEN WE'LL FIND THE AREA OF THIS LARGER BLUE TRIANGLE,
AND BECAUSE THE FUNCTION IS NON-NEGATIVE ON THIS INTERVAL
THIS AREA WOULD BE POSITIVE.
SO THE DEFINITE INTEGRAL FROM 0 TO 6 OF 3/2X - 3
IS GOING TO BE EQUAL TO THE OPPOSITE OF THE RED AREA
+ THE BLUE AREA.
LET'S CALL THIS LENGTH THE BASE 2 AND THIS LENGTH THE HEIGHT 3,
AND FOR THE LARGE BLUE TRIANGLE FROM 2 TO 6 THE BASE WOULD BE 4,
AND THE HEIGHT WOULD BE 6.
BUT AGAIN, THIS FIRST AREA IS GOING TO BE NEGATIVE
BECAUSE THE FUNCTION IS NEGATIVE ON THAT INTERVAL.
SO THE DEFINITE INTEGRAL IS GOING TO BE EQUAL TO
THE OPPOSITE OF 2 x 3 DIVIDED BY 2 + 4 x 6 DIVIDED BY 2.
SO THIS WOULD GIVE US -6 DIVIDED BY 2 IS -3 + 24/2 IS 12.
SO THE DEFINITE INTEGRAL IS EQUAL TO -3 + 12
WHICH IS EQUAL TO 9.
SO EVEN THOUGH A PIECE OF THIS FUNCTION WAS NEGATIVE
ON THE GIVEN CLOSED INTERVAL,
AS LONG AS WE RECOGNIZE THAT FROM 0 TO 2 THE AREA
WOULD BE NEGATIVE BECAUSE THE FUNCTION VALUES ARE NEGATIVE
WE CAN STILL EVALUATE THIS DEFINITE INTEGRAL
USING GEOMETRIC FORMULAS.
I HOPE YOU FOUND THIS EXPLANATION HELPFUL.