Tip:
Highlight text to annotate it
X
- WE WANT TO EVALUATE THE INTEGRALS BELOW
BY INTERPRETING THEM IN TERMS OF AREA.
WELL WE CAN DETERMINE THE VALUE OF A DEFINITE INTERVAL
BY DETERMINING THE AREA BOUNDED BY A FUNCTION IN THE X AXIS
AS LONG AS WE CONSIDER THE AREA ABOVE THE X AXIS
AS A POSITIVE VALUE AND THE AREA BELOW THE X AXIS
AS A NEGATIVE VALUE.
SO FIRST WE HAVE THE INTEGRAL OF F FROM 0 TO 2
WHICH WOULD BE OVER THIS INTERVAL HERE.
NOTICE HOW OVER THIS INTERVAL THE AREA BOUNDED BY THE FUNCTION
IN THE X AXIS IS ABOVE THE X AXIS
AND THEREFORE THE VALUE OF THIS DEFINITE INTEGRAL
WILL BE EQUAL TO THIS AREA AND IT WILL BE POSITIVE.
LET'S GO AHEAD AND CALL THIS AREA "A" SUB 1.
NOTICE HOW IT'S A TRAPEZOID
SO WE COULD USE THE TRAPEZOID FORMULA
TO FIND THE AREA OF THIS REGION
OR WE COULD ALSO CUT THIS INTO A RECTANGLE AND A TRIANGLE
OR BECAUSE OF THE SHAPE WE CAN ACTUALLY
JUST COUNT THE NUMBER OF SQUARES.
NOTICE HOW WE HAVE THREE FULL SQUARES AND TWO HALF SQUARES
SO THE AREA IS 4 SQUARE UNITS.
SO WE CAN SAY "A" SUB 1 IS EQUAL TO 4
AND THEREFORE THE INTEGRAL OF F FROM 0 TO 2 IS EQUAL TO 4 AGAIN
BECAUSE THE AREA IS ABOVE THE X AXIS.
NEXT, WE HAVE THE INTEGRAL OF F FROM 2 TO 7.
SO NOW WE'RE INTEGRATING OVER THIS INTERVAL HERE.
SO NOTICE IN THIS CASE, WE HAVE TWO AREAS
ONE ABOVE THE X AXIS AND ONE BELOW THE X AXIS.
LET'S BEGIN BY DETERMINING THE AREA OF EACH OF THESE REGIONS.
LET'S CALL THIS "A" SUB 2 AND WE'LL CALL THIS "A" SUB 3.
NOTICE "A" SUB 2 IS A RIGHT TRIANGLE
SO WE COULD USE THE FORMULA 1/2 BASE x HEIGHT
OR AGAIN, BECAUSE OF THE SHAPE WE CAN JUST COUNT THE SQUARES.
WE HAVE 3 FULL SQUARES AND 1, 2, 3 HALF SQUARES.
THAT WOULD BE 3 + 1 1/2 OF 4 1/2 OR 4.5 SQUARE UNITS FOR THE AREA
AND THEN FOR "A" SUB 3 WE HAVE A RIGHT TRIANGLE
WITH A BASE AND A HEIGHT OF 2
OR AGAIN WE CAN JUST COUNT THE SQUARES.
WE HAVE 1 FULL SQUARE AND 2 HALF SQUARES
THEREFORE THE AREA WOULD BE 2 SQUARE UNITS.
BUT NOW WHEN IT COMES TO EVALUATING
THE INTEGRAL WE'LL CONSIDER "A" SUB 2 A POSITIVE VALUE
BECAUSE THE AREA IS ABOVE THE X AXIS
AND WE'LL CONSIDER "A" SUB 3 A NEGATIVE VALUE
BECAUSE THE AREA IS BELOW THE X AXIS.
WHICH MEANS THE INTEGRAL OF F FROM 2 TO 7
WOULD BE EQUAL TO THE POSITIVE VALUE OF "A" SUB 2
SO THAT'S 4.5 AND THEN + THE NEGATIVE VALUE OF "A" SUB 3
SO EITHER + -2 OR JUST - 2 GIVING US A VALUE OF 2.5.
FOR OUR LAST INTEGRAL, WE HAVE THE INTEGRAL OF F FROM 0 TO 9
SO NOW WE'RE INTEGRATING OVER THIS ENTIRE INTERVAL HERE.
SO NOTICE NOW FROM 7 TO 9 WE'LL HAVE THIS AREA HERE BELOW
THE X AXIS THAT'S A 2 BY 2 SQUARE.
SO WE'LL CALL THIS "A" SUB 4
AND BECAUSE WE HAVE A 2 BY 2 SQUARE WE'LL SAY
"A" SUB 4 EQUALS 4 SQUARE UNITS.
BUT AGAIN WHEN IT COMES
TO EVALUATING THE DEFINITE INTEGRAL
WE'LL CONSIDER THE POSITIVE VALUE FOR "A" SUB 1
AND "A" SUB 2 AND THE NEGATIVE VALUES OF "A" SUB 3
AND "A" SUB 4.
WHICH MEANS THE INTEGRAL IS EQUAL TO
"A" SUB 1 SO 4 + "A" SUB 2 WHICH IS 4.5
AND THEN EITHER + -2 FOR "A" SUB 3 OR JUST - 2 AND THEN + -4
OR JUST - 4 FOR "A" SUB 4.
SO WE HAVE 8.5 - 2 THAT'S 6.5 - 4 WOULD BE 2.5.
I HOPE YOU FOUND THIS HELPFUL.