- LET A OF X REPRESENT THE AREA BOUNDED BY THE GRAPH,
THE HORIZONTAL AXIS, AND THE VERTICAL LINES T = 0 AND T = X
FOR THE GRAPH BELOW.
WE WANT TO EVALUATE A OF X FOR X = 1, 2, 3, AND 4.
SO FOR A OF 1 WE'RE GOING TO DETERMINE THE AREA
BOUNDED BY T = 0 AND T = 1 WHICH SHOULD BE THE AREA LABELED
A SUB 1.
BECAUSE A SUB 1 IS A TRAPEZOID
WE'LL USE THE AREA FORMULA AREA = 1/2 x THE QUANTITY B SUB 1 + B
SUB 2 x H WHERE B SUB 1 AND B SUB 2
ARE THE TWO BASES OR THE LENGTH OF THE TWO PARALLEL SIDES
AND H WOULD BE THE HEIGHT.
SO LOOKING AT A SUB 1 NOTICE HOW WE CALL THIS B SUB 1.
IT HAS LENGTH 7.
THEY CALL THIS B SUB 2.
IT HAS LENGTH 5.
AND THE HEIGHT WOULD THIS LENGTH HERE WHICH IS 1 UNIT.
SO A OF 1 WOULD BE EQUAL TO 1/2 x THE QUANTITY (7 + 5)
x THE HEIGHT OF 1.
THIS WOULD JUST BE 1/2 x 12.
1/2 x 12 IS 6,
SO A OF 1 IS 6 SQUARE UNITS.
NOW FOR A OF 2 WE WANT THE AREA BELOW THE FUNCTION,
ABOVE THE HORIZONTAL AXIS BETWEEN T = 0 AND T = 2
WHICH WOULD BE THE AREA A SUB 1 + A SUB 2,
BUT WE ALREADY KNOW THAT A SUB 1 IS 6 + A SUB 2.
WELL, A SUB 2 IS A RECTANGLE,
SO WE CAN FIND THE AREA USING THE FORMULA
LENGTH x WIDTH WHERE THE LENGTH WOULD BE 5 UNITS,
AND THE WIDTH WOULD BE 1 UNIT.
AND THEREFORE THE AREA IS 5 x 1 SQUARE UNITS
WHICH WOULD JUST BE 11 SQUARE UNITS.
NOW FOR A SUB 3 WE WANT THE AREA BELOW THE FUNCTION,
ABOVE THE HORIZONTAL AXIS BETWEEN T = 0 AND T = 3
WHICH WOULD BE A SUB 1 + A SUB 2 + A SUB 3.
WELL, WE ALREADY KNOW THAT A SUB 1 + A SUB 2 IS 11,
SO WE HAVE 11 + THE AREA OF THE TRAPEZOID LABELED A SUB 3
WHERE ONE BASE HAS A LENGTH OF 5 UNITS,
THE OTHER BASE HAS A LENGTH OF 2 UNITS, AND THE HEIGHT IS 1 UNIT.
SO WE'D HAVE 1/2 x THE QUANTITY (5 + 2) x 1.
THIS IS GOING TO BE 11 + (1/2 x 7) WHICH WOULD BE 11 + 7/2.
WELL, 7/2 IS 3 1/2 OR 3.5, SO A OF 3 IS 14.5,
THIS WOULD BE SQUARE UNITS.
AND FINALLY FOR A OF 4 WHICH,
AGAIN, IS THE AREA BELOW THE FUNCTION,
ABOVE THE HORIZONTAL AXIS BETWEEN T = 0 AND T = 4,
WHICH MEANS THIS WOULD BE A SUB 1 + A SUB 2 + A SUB 3 + A SUB 4,
BUT WE ALREADY KNOW THAT A SUB 1 + A SUB 2 + A SUB 3 IS 14.5,
AND A SUB 4 IS ANOTHER TRAPEZOID WHERE ONE BASE IS 2 UNITS,
THE OTHER BASE IS 4 UNITS, AND THE HEIGHT IS 1.
SO WE'D HAVE 1/2 x (2 + 4) x 1.
SO WE HAVE 14.5 + THIS WOULD JUST BE 1/2 x 6
WHICH WOULD BE 3, SO WE'LL HAVE 14.5 + 3
WHICH GIVES US 17.5 SQUARE UNITS.
I HOPE YOU FOUND THIS HELPFUL.