- WELCOME.
TODAY MARKS THE SHIFT FROM OUR STUDY OF THE DERIVATIVE
AND THE TANGENT LINE
TO THE AREA UNDER A GRAPH.
THE GOALS OF THIS VIDEO ARE TO DETERMINE THE AREA UNDER A GRAPH
AND ALSO TO APPROXIMATE THE AREA UNDER A GRAPH USING RECTANGLES.
IN MANY CASES, IT'S USEFUL AND MEANINGFUL
TO FIND THE AREA UNDER A GRAPH.
FOR EXAMPLE, IF A CAR IS TRAVELING 65 MILES PER HOUR
FOR 3 HOURS
THE GRAPH WOULD LOOK LIKE THIS.
THE AREA UNDER THE GRAPH AND ABOVE THE X-AXIS
WOULD BE THE SHADED REGION.
IF WE FIND THE AREA OF THIS, WHICH IS A RECTANGLE,
WHICH WOULD BE LENGTH x WIDTH OR 65 x 3,
WE CAN SEE BELOW HERE THAT THE UNITS OF HOURS SIMPLIFY OUT,
AND WE'RE LEFT WITH A TOTAL OF 195 MILES.
SO THIS AREA REPRESENTS THE TOTAL DISTANCE TRAVELED
IN THIS CASE.
HERE'S ANOTHER EXAMPLE.
THIS GRAPH REPRESENTS THE MANUFACTURING COST PER PHONE
FOR A CERTAIN NUMBER OF CELL PHONES.
IF WE TAKE A LOOK AT THE AREA UNDER THIS GRAPH
AND ABOVE THE X-AXIS
WE COULD DIVIDE IT UP INTO FOUR DIFFERENT REGIONS
WHERE EACH REGION WOULD BE A RECTANGLE
AND IT WOULD LOOK LIKE THIS.
SO AGAIN, TO FIND THE AREA OF EACH RECTANGLE
WE'D MULTIPLY THE LENGTH x THE WIDTH.
SO IN THIS CASE THE YELLOW AREA WOULD BE 160 x 2,000.
THIS AREA WOULD BE 100 x 3,000.
THIS AREA WOULD BE 60 x 2,000, AND THEN LASTLY 20 x 3,000.
SO THIS TOTAL AREA,
BECAUSE WE'RE MULTIPLYING THE COST PER PHONE
x THE NUMBER OF CELL PHONES,
WOULD REPRESENT THE TOTAL COST OF PRODUCING 10,000 PHONES.
NOW, WE COULD USE THIS INFORMATION TO DIVIDE BY 10,000
AND DETERMINE THAT THE COST PER PHONE IS $80,
ALL THIS GATHERED BY DETERMINING THE AREA UNDER THE GRAPH.
HERE'S ONE MORE EXAMPLE.
SUNSET NURSERY DETERMINES THE COST TO GROW MESQUITE TREES
LARGE ENOUGH TO SELL IN DOLLARS PER TREE
TO BE THIS FUNCTION WHERE X IS THE NUMBER OF TREES.
NOW KEEP IN MIND THIS IS THE COST PER TREE,
SO THIS WOULD BE EQUIVALENT TO MARGINAL COST
OR LARGE C PRIME OF X.
THAT'S THE REASON WHY THEY USED THE LOWER CASE C
FOR THIS COST FUNCTION.
WE WANT TO GRAPH THE FUNCTION, DETERMINE THE AREA,
AND ALSO INTERPRET THE MEANING.
SO LET'S FIRST TAKE A LOOK AT THE THE GRAPH.
IT'S A LINE WITH A NEGATIVE SLOPE.
HERE IT IS.
WE HAVE THE NUMBER OF TREES ON THE X-AXIS,
THE COST PER TREE IN DOLLARS ON THE Y-AXIS.
NOW WHAT'S UNIQUE ABOUT THIS ONE IS THE AREA UNDER THE GRAPH
AND ABOVE THE X-AXIS IS A TRAPEZOID.
AND REMEMBER THAT THE AREA OF A TRAPEZOID
IS GIVEN BY THIS FORMULA.
SO USING THIS GEOMETRIC FORMULA WE CAN DETERMINE THAT THE AREA
WOULD BE 1/2 x BASE 1 + BASE 2.
NOW, IF WE TURN THIS TRAPEZOID UP THIS WOULD BE BASE 1.
THIS WOULD BE BASE 2.
SO BASE 1 WE'LL CALL 50.
SO BASE 2 WOULD BE 10,
AND THIS LENGTH HERE WOULD BE THE HEIGHT WHICH IS 200
WHICH IS EQUAL TO 6,000.
THE AREA OF THIS REGION IS 6,000 UNITS,
BUT MORE IMPORTANTLY WE WANT TO INTERPRET THE MEANING OF THIS.
WE'RE MULTIPLYING THE COST PER TREE IN DOLLARS
TO NUMBER OF TREES,
SO WE CAN CONCLUDE THAT THE TOTAL COST TO GROW 200 TREES
WOULD BE A TOTAL OF $6,000.
NOW THE QUESTION BECOMES HOW DO WE FIND THE AREA UNDER A GRAPH
IF WE CAN'T USE A SIMPLE GEOMETRIC FORMULA.
IF WE CONSIDER THIS GRAPH ON THE INTERVAL FROM 0 TO 8
AND ABOVE THE X-AXIS,
COULD WE APPROXIMATE THE AREA UNDER THE GRAPH?
WELL, ONE THING WE COULD DO
IS DIVIDE THE INTERVAL FROM 0 TO 8 INTO SMALLER INTERVALS
AND THEN MAKE RECTANGLES,
FIND THE AREA OF THESE RECTANGLES,
AND WE COULD SAY THAT IT WOULD BE APPROXIMATELY EQUAL
TO THE AREA UNDER THE GRAPH.
WE CAN SEE THAT THESE TWO RECTANGLES
WOULD BE TOO SMALL OF AN AREA,
BUT THESE TWO OVER HERE WOULD BE TOO LARGE,
AND SO IT WOULD BE VERY CLOSE TO THE AREA UNDER THE GRAPH.
THE MORE RECTANGLES WE USED
THE MORE ACCURATE OUR ESTIMATE WOULD BE.
SO HERE IS THE IDEA.
IF WE LET N EQUAL THE NUMBER OF RECTANGLES
THAT WE ARE GOING TO USE
WHERE A WOULD EQUAL 0 AND B WOULD EQUAL 8,
BASED UPON THE INTERVAL WE ARE CONSIDERING
THE WIDTH OF EACH RECTANGLE WOULD BE DELTA X
WHICH WOULD BE EQUAL TO B - A DIVIDED BY N,
SO IN THIS CASE, EACH RECTANGLE WOULD HAVE A WIDTH OF 2 UNITS.
ONCE WE DIVIDE THIS INTO WIDTHS OF 2
WE DO HAVE TO DECIDE WHETHER WE WANT TO USE THE LEFT SIDE
OR THE RIGHT SIDE
TO REPRESENT THE HEIGHT OF THE RECTANGLE,
AND IN MOST CASES IN THIS TEXTBOOK
WE'LL USE THE LEFT SIDE OF THE INTERVAL
TO DETERMINE THE HEIGHT OF OUR RECTANGLE.
LET'S TAKE A LOOK AT THIS MORE CLOSELY.
IF I CALL THE AREA OF THIS RECTANGLE AREA 1, AREA 2,
AREA 3, AND SO ON,
WE DO NEED JUST TO DETERMINE THE LENGTH x THE WIDTH.
THIS FIRST RECTANGLE, F OF 0 WOULD BE THE HEIGHT
AND 2 WOULD BE THE WIDTH.
IN THE SECOND RECTANGLE F OF 2 WOULD BE THE HEIGHT,
AND 2, OF COURSE, IS OUR WIDTH.
THE THIRD RECTANGLE, F OF 4,
AND THE FOURTH RECTANGLE WOULD HAVE A LENGTH OF F OF 6.
SO WE COULD SAY OUR AREA WOULD BE APPROXIMATELY EQUAL
TO F OF 0 x 2.
THAT'S THE AREA OF THE FIRST RECTANGLE + F OF 2 x 2
+ F OF 4 x 2 + F OF 6 x 2.
NOW REMEMBER TO FIND THESE HEIGHTS
WE HAVE TO SUBSTITUTE THESE VALUES
INTO THE ORIGINAL FUNCTION HERE.
ONE WAY TO DO THIS WOULD BE TO USE THE GRAPHING CALCULATOR
TO SAVE OURSELVES SOME TIME IN THE CALCULATIONS,
SO IF WE TYPE OUR FUNCTION INTO Y1 AND GO TO OUR TABLE
WE CAN JUST TYPE IN 0, 2, 4, AND 6 FOR OUR VALUES OF X,
AND THESE WOULD BE OUR FUNCTION VALUES
WE NEED TO FIND THESE AREAS.
YOU CAN SEE I'VE ALREADY DONE THAT,
SO LET'S GO AHEAD AND GO BACK TO THE SCREEN AND WRITE THIS OUT.
THIS COMES OUT TO APPROXIMATELY 62.8 SQUARE UNITS.
NOW LOOKING AT THE SAME PROBLEM,
IF I WOULD HAVE USED THE RIGHT SIDE OF THE INTERVAL
TO DETERMINE THE HEIGHT OF EACH RECTANGLE,
YOU CAN SEE THE PICTURE WOULD LOOK DIFFERENT,
BUT IT STILL WOULD BE A GOOD APPROXIMATION OF THE AREA
UNDER THE GRAPH.
LET'S TAKE A LOOK AT THIS PROBLEM IN ANOTHER WAY.
THIS IS THE AREA THAT WE ACTUALLY WANT TO FIND
UNDER OUR FUNCTION,
BUT SINCE WE DON'T HAVE THE ABILITY TO DO THAT
WHAT WE'RE DOING IS USING RECTANGLES
TO APPROXIMATE THE AREA UNDER THE GRAPH.
AS WE INCREASE THE NUMBER OF RECTANGLES,
YOU CAN SEE OUR APPROXIMATION GETS BETTER AND BETTER.
AND YOU CAN KIND OF GET THE IDEA IF N WENT TO INFINITY
IT ACTUALLY WOULD APPROACH THE AREA UNDER THE FUNCTION.
NOW THESE ARE THE LEFT-SIDED RECTANGLES
BECAUSE I'M USING THE LEFT OF EACH INTERVAL TO FIND THAT AREA,
BUT OF COURSE WE COULD USE THE RIGHT
AND THE SAME IDEA WOULD DEVELOP.
OKAY. LET'S GO BACK TO OUR PRESENTATION.
SO AGAIN, IT DOESN'T MATTER
WHICH SIDE OF THE INTERVAL YOU USE,
BUT THE MORE RECTANGLES THE BETTER.
AND SO THIS IS OUR SIGMA NOTATION
OR OUR SUMMATION NOTATION
THAT REPRESENTS THE AREA OF EACH RECTANGLE
BASED UPON HOW MANY RECTANGLES WE USE.
SO THE DEFINITION OF AREA OF A REGION IN THE PLANE
WOULD BE THE LIMIT AS N APPROACHES +INFINITY
OF THIS SUMMATION
WHICH REPRESENTS THE AREA OF ALL THOSE RECTANGLES.
OKAY. LET'S TRY ONE MORE EXAMPLE.
APPROXIMATE THE AREA UNDER THE FUNCTION F OF X
IS EQUAL TO 1 OVER X SQUARED ON THE INTERVAL FROM 1 TO 4
USING 3 AND THEN 6 RECTANGLES.
REMEMBER N REPRESENTS THE NUMBER OF RECTANGLES.
SO TO FIND THE WIDTH OF EACH INTERVAL
WE'LL CALCULATE OUR DELTA X WHICH WOULD BE B - A,
SO 4 - 1 DIVIDED BY 3 OF COURSE WOULD BE 1.
SO WE'LL DIVIDE THE INTERVAL INTO SUBINTERVALS OF WIDTH 1,
AND IN THIS PROBLEM WE'LL USE THE LEFT SIDE OF EACH INTERVAL.
SO THIS FIRST INTERVAL,
THIS WILL BE THE HEIGHT OF OUR FIRST RECTANGLE.
WE'LL CALL THIS AREA 1.
ON THE SECOND INTERVAL WE'LL USE THE LEFT SIDE.
THAT WOULD BE OUR AREA 2,
AND THEN THE THIRD INTERVAL, THIS WOULD BE AREA 3.
SO WE CAN APPROXIMATE THIS AREA BY FINDING F OF--
REMEMBER THE FIRST HEIGHT WOULD BE (F OF 1 x 2) + (F OF 2 x 2)
+ (F OF 3 x 2).
SO NOW WE NEED TO EVALUATE THE FUNCTION AT X = 1, 2, AND 3.
THAT WOULD LOOK LIKE THIS.
TO SAVE SOME TIME I HAVE ALREADY DONE THIS,
AND IT COMES OUT TO 49/36
WHICH IS APPROXIMATELY EQUAL TO 1.36.
THAT WOULD BE THE APPROXIMATE AREA UNDER THIS FUNCTION
IF WE USED 3 RECTANGLES AND USED THE LEFT SIDE OF THE INTERVAL.
THE LAST EXAMPLE WE WANT TO DO THE SAME THING,
BUT NOW WE WANT TO HAVE 6 RECTANGLES.
SO N IS EQUAL IS 6.
SO OUR DELTA X WILL EQUAL 4 - 1 DIVIDED BY 6 WHICH IS 0.5,
SO WE NEED TO DIVIDE THE INTERVAL FROM 1 TO 4
INTO 6 EQUAL INTERVALS OF WIDTH 0.5,
AND THAT WOULD LOOK THIS.
LET'S GO AHEAD AND SKETCH IN OUR RECTANGLES.
AGAIN, WE'LL USE THE LEFT SIDE, SO THERE'S OUR FIRST RECTANGLE.
THERE'S OUR SECOND, THIRD, FOURTH, FIFTH, AND SIXTH.
NOTICE HOW THESE RECTANGLES
ARE LARGER THAN THE AREA UNDER THE FUNCTION.
SOMETIMES THAT'S CALLED THE UPPER SUM.
OKAY, SO OUR AREA WILL CONSIST OF SIX DIFFERENT RECTANGLES.
SO USING OUR GRAPH WE SHOULD BE ABLE TO FIND THE AREA
OF EACH OF THESE RECTANGLES.
LET'S GO AHEAD AND -- IT OUT.
IT WOULD BE F OF 1 x 0.5 AND SO ON.
AGAIN, WE'D HAVE TO SUB ALL THESE X VALUES
INTO THE ORIGINAL FUNCTION
IN ORDER TO FIND THE HEIGHT OF EACH RECTANGLE,
AND TO SAVE SOME TIME I'VE ALREADY DONE THAT
AND CALCULATED ALL THIS OUT,
AND IT COMES OUT TO APPROXIMATELY 1.024.
SINCE WE USED SIX RECTANGLES TO APPROXIMATE THIS AREA
WE COULD CONCLUDE THAT THIS WOULD BE MORE ACCURATE
THAN THE PREVIOUS EXAMPLE
WHERE WE HAD AN APPROXIMATE AREA OF 1.36 USING THREE RECTANGLES.
I HIGHLY SUGGEST SKETCHING OUT YOUR INTERVALS,
SKETCHING YOUR RECTANGLES,
AND THEN FIND THE AREA OF EACH ONE TO APPROXIMATE THE AREA.
I HOPE THAT HELPED.
THANK YOU AND HAVE A GOOD DAY.