 – WELCOME TO TWO EXAMPLES INVOLVING THE ADDITION PROPERTY
OF DEFINITE INTEGRALS. TO HELP ILLUSTRATE
THESE EXAMPLES WE’LL ASSUME F OF X IS THIS
FUNCTION HERE, GRAPHED IN RED, WHICH IS A NONNEGATIVE FUNCTION. WE’RE GIVEN THE DEFINITE
INTEGRAL OF F OF X FROM 1 TO 6=THE DEFINITE INTEGRAL OF F
OF X FROM 1 TO 4 + THE DEFINITE INTEGRAL OF F
OF X FROM “A” TO B, AND WE’RE ASKED TO FIND
“A” AND B. BECAUSE F OF X IS NONNEGATIVE, IF WE INTEGRATE F OF X
FROM 1 TO 6 IT WOULD GIVE US THE AREA
OF THIS BLUE SHADED REGION, WHICH MEANS THE SUM OF THESE TWO
DEFINITE INTEGRALS MUST ALSO GIVE US THE SAME AREA. SO LOOKING AT THE RIGHT SIDE
OF THIS EQUATION WE HAVE THE INTEGRAL
OF F OF X FROM 1 TO 4. WELL, IF WE INTEGRATE
THIS FUNCTION ON THE INTERVAL FROM 1 TO 4,
IT WOULD GIVE US THIS AREA HERE. AND, THEREFORE,
TO GET THE TOTAL AREA, THIS SECOND INTEGRAL HERE
MUST BE INTEGRATED ON THE INTERVAL FROM 4 TO 6,
GIVING US THE REMAINING AREA, OR THIS AREA HERE. AND THAT MEANS “A” MUST BE EQUAL
TO 4, AND B MUST BE EQUAL TO 6. LET’S TAKE A LOOK
AT A SECOND EXAMPLE. HERE WE HAVE THE INTEGRAL
OF F OF X FROM 1 TO 3 MUST EQUAL THE DEFINITE INTEGRAL
OF F OF X FROM 1 TO 5 + THE DEFINITE INTEGRAL OF F
OF X FROM “A” TO B. AND ONCE AGAIN, WE WANT TO FIND
THE VALUE OF “A” AND B. SO NOTICE IN THIS CASE
ON THE LEFT SIDE, IF WE INTEGRATE FROM 1 TO 3 WE WOULD GET THIS AREA
HERE IN BLUE, SO WE WANT THE SUM OF THESE TWO
INTEGRALS TO BE EQUAL TO THIS BLUE AREA. BUT NOTICE THIS FIRST DEFINITE
INTEGRAL FROM 1 TO 5 WOULD GIVE US MORE AREA
THAN WE HAVE HERE IN BLUE. IT WOULD ACTUALLY GIVE US
ALL OF THIS AREA HERE, WHICH MEANS YOU NOW WANT TO
SUBTRACT OUT THIS AREA HERE. BUT NOTICE HOW WE DON’T HAVE
A SUBTRACTION SIGN HERE, WE HAVE AN ADDITION SIGN. SO IF WE WANT TO SUBTRACT
OUT THIS AREA HERE, INSTEAD OF INTEGRATING
FROM 3 TO 5, WE CAN CHANGE THE ORDER
OF INTEGRATION AND INTEGRATE FROM 5 TO 3. IF WE INTEGRATE F OF X
FROM 5 TO 3, IT WOULD RETURN
A NEGATIVE VALUE, OR WE CAN THINK OF IT AS
SUBTRACTING OUT THIS AREA HERE, WHICH MEANS “A”
WOULD BE EQUAL TO 5, AND B WOULD BE EQUAL TO 3. THAT’S GOING TO DO IT
FOR THESE TWO EXAMPLES. IN THE NEXT EXAMPLE WE’LL TAKE
A LOOK AT A PROBLEM WHERE WE HAVE A SUBTRACTION
OF TWO DEFINITE INTEGRALS.