Tip:
Highlight text to annotate it
X
- SUPPOSE BIG F OF X HAS A DERIVATIVE LITTLE F OF X,
GRAPHED BELOW, AND BIG F OF 0 = 3.
WE WANT TO FIND THE VALUES OF BIG F OF 2 AND BIG F OF 7.
WELL, BECAUSE BIG F HAS THE DERIVATIVE LITTLE F,
THAT MEANS BIG F IS THE ANTIDERIVATIVE OF LITTLE F,
AND THEREFORE, WE CAN FIND
THE VALUE OF BIG F OF 2 AND BIG F OF 7
USING THE FUNDAMENTAL THEOREM OF CALCULUS,
WHERE THE INTEGRAL OF LITTLE
F FROM "A" TO B = BIG F OF B - BIG F OF "A."
SO, FOR EXAMPLE, TO FIND BIG F OF 2,
WE CAN USE THE INTEGRAL OF LITTLE F FROM 0 TO 2
MUST = BIG F OF 2 - BIG F OF 0.
WE CAN DETERMINE THE VALUE OF THIS DEFINITE INTEGRAL
USING THE GRAPH OF F OF X HERE, AND WE'RE GIVEN BIG F OF 0 = 3.
SO FOR THE NEXT STEP LET'S GO AHEAD AND SOLVE THIS
FOR BIG F OF 2 BY ADDING BIG F OF 0 TO BOTH SIDES.
WE'D HAVE THE INTEGRAL OF LITTLE
F FROM 0 TO 2 + BIG F OF 0 = F OF 2.
AND NOW TO DETERMINE THE VALUE OF THIS DEFINITE INTEGRAL,
WE CAN DETERMINE THE AREA BOUNDED BY THE FUNCTION
AND THE X AXIS OVER THE CLOSED INTERVAL FROM ZERO TO 2.
BUT IF THE AREA IS ABOVE THE X AXIS,
THE VALUE WOULD BE POSITIVE.
IF THE AREA IS BELOW THE X AXIS, THE VALUE WOULD BE NEGATIVE.
SO LOOKING AT THE AREA BOUNDED BY THIS FUNCTION
ON THE CLOSED INTERVAL FROM ZERO TO 2,
WE WANT TO FIND THE AREA OF THIS TRIANGULAR REGION,
BUT BECAUSE THE AREA IS BELOW THE X AXIS,
OR BECAUSE THE FUNCTION VALUES ARE NEGATIVE,
WE DO WANT TO VIEW THIS AS A NEGATIVE VALUE.
SO USING THE AREA FORMULA FOR A TRIANGLE,
AREA = 1/2 BASE x HEIGHT.
WE WOULD HAVE AREA = 1/2 x THE BASE OF 2 x THE HEIGHT OF 3.
SO THE AREA OF THIS TRIANGLE IS EQUAL TO 3 SQUARE UNITS,
BUT BECAUSE THE AREA IS BELOW THE X AXIS,
THE VALUE OF THIS DEFINITE INTEGRAL IS GOING TO BE -3,
AND THEN + THE VALUE OF BIG F OF 0, WHICH IS 3.
SO NOTICE THAT BIG F OF 2 = 0.
NOW LET'S DETERMINE THE VALUE OF BIG F OF 7.
THE DEFINITE INTEGRAL OF LITTLE
F FROM 0 TO 7 MUST = BIG F OF 7 - BIG F OF 0.
TO SOLVE THIS FOR BIG F OF 7
WE'LL ADD BIG F OF 0 TO BOTH SIDES,
GIVING US THE INTEGRAL OF LITTLE
F FROM 0 TO 7 + BIG F OF 0 = BIG F OF 7.
AND NOW WE'LL WORK ON DETERMINING
THE VALUE OF THIS DEFINITE INTEGRAL
BY ANALYZING THE GRAPH OF OUR FUNCTION F OF X
HERE ON THE SIDE.
LET'S FIRST SHADE THE AREA BOUNDED BY THE FUNCTION
AND THE X AXIS OVER THE CLOSED INTERVAL FROM 0 TO 7.
AND, AGAIN, BECAUSE THIS AREA IS BELOW THE X AXIS
FOR THE VALUE OF THE DEFINITE INTEGRAL,
WE'RE GOING TO VIEW THIS AS A NEGATIVE VALUE.
AND BECAUSE THIS AREA IS ABOVE THE X AXIS,
WE CAN THINK OF THIS AS A POSITIVE VALUE.
SO LET'S GO AHEAD AND CALL THIS "A" SUB 1.
LET'S DIVIDE THIS HERE AND CALL THIS "A" SUB 2 AND "A" SUB 3.
SO THE AREA OF "A" SUB 1 IS GOING TO BE EQUAL TO 1/2
x THE BASE OF 3 x A HEIGHT OF 3.
SO THAT'S EQUAL TO 9/2,
EVEN THOUGH FOR THE DEFINITE INTEGRAL
IT WOULD BE A NEGATIVE VALUE.
AND NOW FOR "A" SUB 2, THIS RIGHT TRIANGLE HERE,
THE AREA IS GOING TO BE EQUAL TO 1/2 x A BASE OF 3,
AND ALSO THE HEIGHT IS 3.
SO THIS IS ALSO 9/2,
BUT THIS VALUE FOR THE DEFINITE INTEGRAL
IS GOING TO BE POSITIVE BECAUSE THE AREA IS ABOVE THE X AXIS.
AND THEN FOR AREA SUB 3,
WE CAN SEE THE AREA IS GOING TO BE 3 SQUARE UNITS.
SO THE VALUE OF THIS DEFINITE INTEGRAL
IS GOING TO BE EQUAL TO "-A" SUB 1, OR -9/2 + "A" SUB 2,
WHICH IS 9/2, + "A" SUB 3, WHICH IS 3.
AND, AGAIN, THIS VALUE IS NEGATIVE
BECAUSE THE AREA IS BELOW THE X AXIS.
THESE TWO VALUES ARE POSITIVE
BECAUSE THE AREA IS ABOVE THE X AXIS.
AND THEN + BIG F OF 0, WHICH WE'RE GIVEN AS 3 = BIG F OF 7.
SO THESE TWO SIMPLIFY OUT, AND SO WE HAVE 3 + 3 = BIG F OF 7,
SO BIG F OF 7 = 6.
I HOPE YOU'VE FOUND THIS HELPFUL.