- WELCOME TO AN EXAMPLE
OF THE SECOND FUNDAMENTAL THEOREM OF CALCULUS
THAT ALSO REQUIRES A CHAIN RULE.
HERE WE'RE GIVEN F OF X EQUALS THIS INTEGRAL,
AND ASKED TO FIND F PRIME OF X AND F PRIME OF 1.
SO ONE APPROACH WOULD BE TO FIND F OF X BY INTEGRATING,
AND THEN FIND F PRIME OF X BY DIFFERENTIATING.
BUT WE CAN ALSO FIND F PRIME OF X
BY USING 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 = F OF X.
SO A COUPLE THINGS TO NOTICE HERE.
THE LOWER LIMIT OF INTEGRATION IS OUR CONSTANT "A. "
THE UPPER LIMIT OF INTEGRATION IS JUST X,
WHICH IS THE SAME VARIABLE
IN WHICH WE DIFFERENTIATE WITH RESPECT TO.
SO TO FIND THE DERIVATIVE OF THIS INTEGRAL
WE SUBSTITUTE X FOR T GIVING US F OF X.
SO IF F OF X IS EQUAL TO THIS INTEGRAL,
WE CAN SAY F PRIME OF X WOULD BE EQUAL TO THE DERIVATIVE
WITH RESPECT TO X OF THE GIVEN INTEGRAL.
BUT IF WE COMPARE THE FORM OF THIS INTEGRAL
TO THE INTEGRAL IN OUR THEOREM,
WE DO HAVE A SLIGHT PROBLEM.
NOTICE IN THE THEOREM THE UPPER LIMIT OF INTEGRATION IS JUST X,
AND WE HAVE X TO THE FOURTH.
THIS MEANS WE'LL HAVE TO PERFORM A U-SUBSTITUTION
BEFORE WE APPLY THE SECOND FUNDAMENTAL THEOREM OF CALCULUS.
MEANING IF WE LET U = X TO THE FOURTH,
THEN WE COULD SAY THAT F PRIME OF U, NOT F PRIME OF X,
WOULD BE EQUAL TO THE DERIVATIVE WITH RESPECT TO U
OF THE INTEGRAL FROM -1 TO U, NOT X TO THE FOURTH, OF F OF T.
SO NOW WE CAN USE THE THEOREM TO FIND F PRIME OF U,
BUT REMEMBER OUR GOAL IS TO FIND F PRIME OF X,
WHICH MEANS WE'LL HAVE TO APPLY THE CHAIN RULE.
F PRIME OF X WOULD BE EQUAL TO F PRIME OF U x U PRIME OR DFDU
x DUDX.
SO REMEMBER WE LET U = X TO THE FOURTH,
WHICH MEANS U PRIME, OR DUDX,
WOULD BE EQUAL TO 4X TO THE THIRD.
AND NOW WE CAN FIND F PRIME OF U
BY APPLYING THE SECOND FUNDAMENTAL THEOREM OF CALCULUS.
IN ORDER TO DO THIS WE'LL SUBSTITUTE U FOR T.
SO F PRIME OF U WOULD BE EQUAL TO U TO THE THIRD.
SO F PRIME OF X IS EQUAL TO DFDU OR F PRIME OF U,
WHICH WOULD BE U TO THE THIRD
x DUDX, WHICH IS 4X TO THE THIRD.
BUT, OF COURSE, WE WANT F PRIME OF X IN TERMS OF X,
SO NOW WE'LL SUBSTITUTE X TO THE FOURTH FOR U.
SO WE'D HAVE X TO THE FOURTH RAISED TO THE THIRD
x 4X TO THE THIRD.
WELL, THIS WOULD BE X TO THE TWELFTH x 4X TO THE THIRD,
WHICH WOULD BE 4X TO THE FIFTEENTH,
WHICH AGAIN IS F PRIME OF X, WHICH WE WERE LOOKING FOR.
SO THIS IS 4X TO THE FIFTEENTH POWER.
SO TO FIND F PRIME OF 1, WE'LL SUBSTITUTE 1 FOR X.
OF COURSE, 1 TO THE FIFTEENTH IS JUST 1,
SO WE HAVE 4 x 1 OR 4.
SO WHILE WE ARE DONE,
LET'S GO AHEAD AND CHECK THIS BY INTEGRATING TO FIND F OF X
AND THEN DIFFERENTIATING TO CHECK F PRIME OF X.
AND LET'S DO THIS ON THE NEXT SLIDE.
SO THE ANTIDERIVATIVE FUNCTION WOULD BE T TO THE FOURTH
DIVIDED BY 4,
WHICH WE'D EVALUATE AT X TO THE FOURTH, AND THEN -1.
SO WHEN T = X TO THE FOURTH,
WE'LL HAVE X TO THE FOURTH TO THE FOURTH DIVIDED BY 4.
AND THEN WHEN T IS -1,
WE'LL HAVE -1 TO THE FOURTH DIVIDED BY 4.
SO WE CAN SAY F OF X IS EQUAL TO,
THIS WOULD BE X TO THE SIXTEENTH DIVIDED BY 4
OR 1/4X TO THE SIXTEENTH,
AND THIS WOULD BE - 1/4.
AND NOW WE CAN FIND F PRIME OF X WOULD BE EQUAL
TO 1/4 x 16X TO THE FIFTEENTH - 0.
HERE WE HAVE A COMMON FACTOR OF 4.
THERE'S 1, 4, AND 4, AND FOUR 4s IN SIXTEEN.
SO F PRIME OF X = 4X TO THE FIFTEENTH,
WHICH IS THE SAME THING WE FOUND
USING THE SECOND FUNDAMENTAL THEOREM OF CALCULUS,
AS WE SEE HERE.
I HOPE YOU HAVE FOUND THIS HELPFUL.