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X
- IF F OF X = THE INTEGRAL OF 6T TO THE SECOND + 3T - 2
FROM 0 TO X,
THEN WE WANT TO FIND F PRIME OF X, F PRIME OF 2,
AND F DOUBLE PRIME.
SO ONE WAY OF APPROACHING THIS
WOULD BE TO EVALUATE THIS INTEGRAL
AND THEN FIND THE DERIVATIVE OF THE RESULT.
BUT WE CAN ALSO DETERMINE F PRIME OF X
BY APPLYING THE SECOND FUNDAMENTAL THEOREM OF CALCULUS
STATED HERE.
WHERE IF F IS CONTINUOUS ON AN OPEN INTERVAL
CONTAINING A CONSTANT "A,"
THEN FOR EVERY X IN THE INTERVAL
THE DERIVATIVE WITH RESPECT TO X OF THE INTEGRAL OF F OF T
FROM "A" TO X IS EQUAL TO F OF X.
SO LOOKING AT THE INTEGRAL
NOTICE HOW THE LOWER LIMIT OF INTEGRATION IS A CONSTANT "A,"
WHERE "A" IS IN AN INTERVAL WHERE F IS CONTINUOUS.
THE UPPER LIMIT OF INTEGRATION IS X,
WHICH IS THE SAME VARIABLE
IN WHICH WE'RE FINDING THE DERIVATIVE WITH RESPECT TO.
SO TO FIND THE DERIVATIVE OF THIS INTEGRAL
WE SIMPLY SUBSTITUTE X FOR T INTO F.
SO GOING BACK TO OUR EXAMPLE, IF THE INTEGRAL IS EQUAL TO F,
WE CAN SAY THAT F PRIME OF X WOULD BE EQUAL TO THE DERIVATIVE
WITH RESPECT TO X OF THE INTEGRAL.
NOTICE HOW THIS INTEGRAL DOES FIT THE FORM IN OUR THEOREM,
AND, THEREFORE, TO FIND THE DERIVATIVE OF THIS INTEGRAL
WE SUBSTITUTE X FOR T,
WHICH WOULD JUST GIVE US 6X TO THE SECOND + 3X - 2.
SO, AGAIN, F PRIME OF X = 6X TO THE SECOND + 3X - 2,
AND, THEREFORE, F PRIME OF 2 WE'LL SUBSTITUTE 2 FOR X.
THAT'LL BE 6 x 2 SQUARED + 3 x 2 - 2.
WE'D HAVE 4 x 6, THAT'S 24 + 6, THAT'S 30 - 2 = 28.
AND NOW TO FIND F DOUBLE PRIME,
WE'LL FIND THE DERIVATIVE OF THE FIRST DERIVATIVE,
WHICH WOULD BE 12X + 3,
APPLYING THE POWER RULE OF DIFFERENTIATION.
BUT LET'S ALSO TAKE THE TIME TO FIND F OF X
BY EVALUATING THIS INTEGRAL,
AND THEN WE'LL FIND THE DERIVATIVE
TO VERIFY F PRIME OF X IS CORRECT.
AND LET'S GO AHEAD AND DO THIS ON THE NEXT SLIDE.
SO I'LL FIRST FIND THE ANTIDERIVATIVE,
WHICH WOULD BE 6 x T TO THE THIRD DIVIDED BY 3
+ 3 x T TO THE SECOND DIVIDED BY 2 - 2T.
WE EVALUATE THIS AT X AND ZERO.
LET'S FIRST SIMPLIFY THIS THOUGH.
THIS WOULD BE 2T TO THE THIRD + 3/2T TO THE SECOND - 2T.
WE'LL FIRST SUBSTITUTE X AND THEN WE'LL SUBSTITUTE ZERO.
SO WHEN T = X, WE JUST HAVE 2X TO THE THIRD
+ 3/2X SQUARED - 2X.
AND WHEN T IS 0 WE JUST HAVE 0.
SO NOW WE KNOW THAT F OF X = 2X TO THE THIRD
+ 3/2X SQUARED - 2X,
AND, THEREFORE, WE CAN NOW FIND F PRIME OF X,
WHICH WOULD JUST BE 2 x 3X TO THE SECOND + 3/2 x 2X - 2 x 1,
OR JUST - 2.
SO NOTICE F PRIME OF X = 6X TO THE SECOND + 3X - 2.
WHICH IS THE SAME AS WHAT WE FOUND
WHEN APPLYING THE SECOND FUNDAMENTAL THEOREM OF CALCULUS,
AS WE SEE HERE.
I HOPE YOU HAVE FOUND THIS HELPFUL.