- WE'RE GIVEN THAT F OF X = 
THE INTEGRAL OF 2T TO THE THIRD
MINUS T FROM 1 TO 3X
AND ASKED TO FIND F PRIME OF X 
AND F PRIME OF 1.
SO WE COULD INTEGRATE TO FIND 
F OF X
AND THEN FIND THE DERIVATIVE 
TO FIND F PRIME OF X.
BUT WE CAN ALSO APPLY THE SECOND 
FUNDAMENTAL THEOREM OF CALCULUS
STATED HERE BELOW.
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.
A COUPLE THINGS TO NOTICE HERE.
OUR LOWER LIMIT OF INTEGRATION 
IS THE CONSTANT "A."
THE UPPER LIMIT OF INTEGRATION 
IS JUST X,
THE SAME VARIABLE IN WHICH 
WE'RE FINDING DERIVATIVE
WITH RESPECT TO.
SO TO FIND THE DERIVATIVE 
OF THE INTEGRAL
WE SIMPLY SUBSTITUTE X FOR T.
SO LOOKING AT OUR PROBLEM, 
IF F OF X EQUALS THIS INTEGRAL,
THEN WE CAN SAY THAT F PRIME 
OF X
WOULD BE EQUAL TO THE DERIVATIVE 
WITH RESPECT TO X
OF OUR GIVEN INTEGRAL.
BUT IF WE COMPARE THIS INTEGRAL 
TO THE INTEGRAL IN THE THEOREM
WE DO HAVE A SLIGHT PROBLEM.
NOTICE HERE THE UPPER LIMIT 
OF INTEGRATION IS JUST X
AND HERE IT'S 3X.
THIS MEANS TO APPLY THIS THEOREM
WE'LL HAVE TO PERFORM 
A U-SUBSTITUTION
AND USE THE CHAIN RULE 
TO FIND F PRIME OF X.
AND WE NEED A VARIABLE HERE 
WITH A COEFFICIENT OF 1,
WHICH MEANS IF WE LET U = 3X,
WE CAN ALSO SAY THAT 
NOT F PRIME OF X,
BUT F PRIME OF U
WOULD BE EQUAL TO THE DERIVATIVE 
WITH RESPECT TO U
OF THE INTEGRAL FROM 1 TO U 
OF OUR FUNCTION F OF T.
SO NOW WE CAN APPLY THE SECOND 
FUNDAMENTAL THEOREM OF CALCULUS
TO FIND F PRIME OF U, 
OR DF TO U,
BUT SINCE OUR GOAL IS TO FIND F 
PRIME OF X USING THE CHAIN RULE,
WE SHOULD RECOGNIZE THAT F PRIME 
OF X = F PRIME OF U x U PRIME,
OR DFDU x DUDX.
SO IF WE LET U = 3X, 
NOTICE THAT DUDX, OR U PRIME,
WOULD JUST BE 3.
SO NOW APPLYING THE SECOND 
FUNDAMENTAL THEOREM OF CALCULUS
TO FIND F PRIME OF U, 
WE SUBSTITUTE U FOR T.
WHICH MEANS F PRIME OF U, 
OR DFDU = 2U TO THE THIRD - U.
AND NOW WE HAVE ALL 
THE INFORMATION WE NEED
TO FIND F PRIME OF X.
F PRIME OF X = DFDU, 
OR F PRIME OF U,
WHICH IS 2U TO THE THIRD - U 
x DUDX, WHICH IS 3.
THIS WOULD GIVE US 
6U TO THE THIRD - 3U.
BUT, OF COURSE, WE WANT F PRIME 
OF X IN TERMS OF X,
SO NOW WE'LL SUBSTITUTE 3X 
FOR U.
SO WE HAVE 6 x 3X TO THE THIRD 
- 3 x 3X.
WELL, THIS WOULD BE 
27X CUBED x 6.
162X TO THE THIRD - 9X.
SO AGAIN, THIS IS F PRIME OF X.
AND, THEREFORE, F PRIME OF 1 
WE'LL SUBSTITUTE 1 FOR X,
SO IT WOULD JUST BE 162 - 9, 
WHICH IS EQUAL TO 153.
SO THE BENEFIT OF USING 
THIS THEOREM IS THAT
WE CAN FIND THIS DERIVATIVE
WITHOUT ACTUALLY FINDING 
THIS INTEGRAL.
BUT BECAUSE WE ARE ABLE 
TO INTEGRATE THIS,
LET'S GO AHEAD AND CHECK THIS 
BY FINDING OUR INTEGRAL,
OR F OF X,
AND THEN DIFFERENTIATING 
TO CHECK F PRIME OF X.
LET'S GO AHEAD AND DO THIS 
ON THE NEXT SLIDE.
SO TO DO THIS 
WE WOULD FIRST FIND
THE ANTIDERIVATIVE FUNCTION,
WHICH WOULD BE 2 x T 
TO THE FOURTH
DIVIDED BY 4 - T TO THE SECOND 
DIVIDED BY 2.
LET'S GO AHEAD AND SIMPLIFY 
FIRST.
THIS WOULD BE 1/2T TO THE FOURTH 
- 1/2T SQUARED.
AND NOW WE'LL EVALUATE THIS 
FIRST AT 3X AND THEN AT 1.
SO WHEN T = 3X WE'LL HAVE 1/2 
x 3X TO THE FOURTH
MINUS 1/2 x 3X TO THE SECOND,
AND THEN WE'LL HAVE - THE 
QUANTITY 1/2 x 1 TO THE FOURTH
MINUS 1/2 x 1 TO THE SECOND.
SO HERE WE'RE GOING TO HAVE 81X 
TO THE FOURTH x 1/2,
81 DIVIDED BY 2X TO THE FOURTH.
HERE WE'LL HAVE 9X SQUARED 
x 1/2, SO -9/2X SQUARED.
HERE WE'LL HAVE 
- THE QUANTITY 1/2 - 1/2,
WHICH WOULD JUST BE 0.
SO HERE'S F OF X, AND NOW WE CAN 
FIND F PRIME OF X.
SO WE HAVE 81 DIVIDED BY 2 
x 4X TO THE THIRD - 9/2 x 2X.
LET'S GO AHEAD AND SIMPLIFY.
COMMON FACTOR OF 2 HERE,
AND A COMMON FACTOR OF 2 HERE.
SO THIS WOULD BE 162X 
TO THE THIRD - 9X.
AND THIS IS THE SAME THING 
WE FOUND FOR F PRIME OF X
USING THE SECOND FUNDAMENTAL 
THEOREM OF CALCULUS.
NORMALLY YOU WOULDN'T DO THIS 
BOTH WAYS,
BUT IT IS A NICE WAY 
TO CHECK OUR WORK.
NOTICE HOW F PRIME OF X 
IS THE SAME AS WE HAVE HERE.
THE BENEFIT OF USING THE SECOND 
FUNDAMENTAL THEOREM OF CALCULUS
IS THAT EVEN IF WE WEREN'T ABLE 
TO INTEGRATE THIS INTEGRAL,
WE CAN STILL FIND 
THE DERIVATIVE OF THE INTEGRAL
USING THIS THEOREM.
I HOPE YOU HAVE FOUND THIS 
HELPFUL.
