- THE TEMPERATURE OF A TOWN T MONTHS AFTER JANUARY
CAN BE ESTIMATED BY THE FUNCTION F OF T = -24 COSINE PI
DIVIDED BY 6T + 58.
WE WANT TO FIND THE AVERAGE VALUE
FROM MONTH ONE TO MONTH FOUR.
WHERE AN AVERAGE FUNCTION VALUE OVER THE CLOSED INTERVAL
FROM A TO B IS EQUAL TO 1 DIVIDED BY THE QUANTITY B - A
x THE DEFINITE INTEGRAL OF F OF X FROM A TO B.
SO IN OUR CASE WE'RE TRYING TO FIND THE AVERAGE VALUE
OVER THE CLOSED INTERVAL FROM 1 TO 4.
WHICH MEANS, THE AVERAGE VALUE OR AVERAGE TEMPERATURE
FROM MONTH 1 TO MONTH 4 WOULD BE Y OF A SUB V = 1
DIVIDED BY 4 - 1 x THE INTEGRAL OF F OF T FROM 1 TO 4.
BEFORE WE FIND THIS VALUE THOUGH,
LETS LOOK AT THE GRAPH OF OUR FUNCTION
OVER THIS CLOSED INTERVAL.
SO HERE'S THE GRAPH OF OUR FUNCTION
AND HERE'S THE CLOSED INTERVAL FROM 1 TO 4.
BECAUSE THE FUNCTION IS NON-NEGATIVE
WE KNOW THE VALUE OF JUST THE DEFINITE INTEGRAL
WOULD GIVE US THE AREA BOUNDED BY THE FUNCTION IN THE X-AXIS,
WHICH SHOULD BE THE AREA OF THIS SHADED REGION HERE.
SO IF WE TAKE THIS AREA AND MULTIPLY BY 1
DIVIDED BY THE QUANTITY B - A,
WHICH WOULD BE THE SAME AS DIVIDING BY THE WIDTH
OF THE INTERVAL, THIS WOULD GIVE US THE AVERAGE VALUE,
WHICH WOULD BE THE FUNCTION VALUE
THAT WOULD GIVE US THE HEIGHT OF THE RECTANGLE
OVER THIS INTERVAL MAYBE SOMEWHERE IN HERE
WHERE THE AREA OF THIS RECTANGLE WOULD HAVE THE SAME AREA
AS THE AREA UNDER THE FUNCTION AND ABOVE THE X-AXIS.
SO THIS Y VALUE HERE OR THIS FUNCTION VALUE
WOULD BE THE AVERAGE VALUE.
SO NOW GOING BACK TO OUR DEFINITE INTEGRAL
LET'S EVALUATE THIS.
WE'D HAVE 1/3 x NOW WE'LL FIND THE ANTI DERIVATIVE HERE,
BUT TO FIND THE ANTI DERIVATIVE OF THIS FIRST TERM
WE DO HAVE TO PERFORM U SUBSTITUTION.
NOTICE HOW WE HAVE COSINE PI DIVIDED BY 6T,
SO U WOULD BE EQUAL TO PI DIVIDED BY 6T,
WHICH MEANS DIFFERENTIALLY U OR DU WOULD BE PI DIVIDED BY 6DT.
SO IF WE SOLVE THIS FOR DT BY MULTIPLYING
BOTH SIDES OF THE EQUATION BY THE RECIPROCAL OF PI/6
WHICH WOULD BE 6/PI, WE'D HAVE 6 DIVIDED BY PI DU = DT.
WHICH MEANS, WHEN INTEGRATING HERE
WE'LL HAVE AN EXTRA FACTOR OF 6 DIVIDED BY PI.
SO WE'D HAVE -24 x 6 DIVIDED BY PI x SINE PI DIVIDED BY 6T.
AND THEN PLUS THE ANTI DERIVATIVE OF 58 WOULD BE 58T.
NOW WE'LL EVALUATE THIS AT T = 4, T = 1,
AND THEN FIND THE DIFFERENCE.
SO THIS WILL GET A LITTLE MESSY.
I'M GOING TO HAVE 1/3 x WHEN T = 4
WE'D HAVE -24 X 6 DIVIDED BY PI SINE PI/6 x 4
THAT BE 4PI/6 OR 2PI/3.
AND THEN WE'D HAVE + 58 x 4.
SO ALL OF THIS MINUS THE VALUE WHEN T IS 1.
SO, AGAIN, WE'D HAVE -24 x 6 DIVIDED BY PI x SINE JUST PI/6,
AND THEN + 58 x 1.
FROM HERE WE'D GO TO THE CALCULATOR AND TYPE THIS IN.
THIS WOULD END UP BEING 1/3 x 157.2226 APPROXIMATELY.
AND MULTIPLYING BY 1/3 OR DIVIDING BY 3
THIS WOULD GIVE US APPROXIMATELY 52.41.
SO PROBABLY THE MOST CHALLENGING THING ABOUT THIS PROBLEM
IS BEING CAREFUL WITH THE U SUBSTITUTION AND THEN TRYING
TO ENTER THIS INTO YOUR CALCULATOR CORRECTLY.
IF WE GO BACK AND TAKE A LOOK AT THE GRAPH AGAIN FOR A MOMENT,
NOTICE HOW THE HEIGHT OF THIS RECTANGLE HERE
THAT HAS THE SAME AREA AS THE AREA OF THE CURVE
OVER THIS INTERVAL WILL BE APPROXIMATELY 52.41.
THAT'S THE VALUE THAT WE FOUND BY DETERMINING
THE AVERAGE VALUE.
I HOPE YOU FOUND THIS HELPFUL.