- WE'RE GIVEN H OF T 
= THE INTEGRAL
OF THE SQUARE ROOT OF THE 
QUANTITY 8 + X TO THE FIFTH
FROM 0 TO T TO THE FOURTH.
WE WANT TO FIND H PRIME OF T, 
OR THE DERIVATIVE OF H OF T.
TO DO THIS WE'LL APPLY 
THE SECOND FUNDAMENTAL THEOREM
OF CALCULUS,
STATED HERE BELOW,
WHERE NOTICE HOW THE DERIVATIVE 
OF THIS INTEGRAL OF F OF T
FROM "A" TO X = F OF X.
WE CAN THINK OF THE DERIVATIVE 
AND INTEGRAL UNDOING EACH OTHER,
AND, THEREFORE, TO EVALUATE THIS 
WE SUBSTITUTE X FOR T
INTO THE INTEGRAND,
GIVING US F OF X.
BUT NOTICE HERE THE LOWER LIMIT 
OF INTEGRATION IS A CONSTANT,
AND THE UPPER LIMIT 
OF INTEGRATION IS JUST X.
SO IF THIS ISN'T JUST X,
LIKE IN OUR CASE WE HAVE 
T TO THE FOURTH,
WE WILL HAVE TO APPLY 
THE CHAIN RULE
TO FIND THE DERIVATIVE 
OF THIS INTEGRAL.
SO BECAUSE OUR UPPER LIMIT OF 
INTEGRATION IS T TO THE FOURTH,
WE'RE GOING TO MAKE 
A SUBSTITUTION HERE
AND LET U = T TO THE FOURTH,
WHICH MEANS U PRIME, OR DUDT,
WOULD BE EQUAL TO 
4T TO THE THIRD.
LET'S WRITE THIS 
IN TERMS OF U NOW.
WE COULD SAY THAT H PRIME OF U
WOULD BE EQUAL TO THE DERIVATIVE 
WITH RESPECT TO U
OF THE INTEGRAL OF THE SQUARE 
ROOT OF THE QUANTITY
8 + X TO THE FIFTH DX,
BUT NOW IT'D BE FROM 0 TO U, 
NOT 0 TO T TO THE FOURTH.
BUT NOTICE IN THIS FORM WE CAN 
DIRECTLY APPLY THE THEOREM
STATED HERE BELOW
AND SUBSTITUTE U INTO THE 
INTEGRAND TO FIND H PRIME OF U.
THIS WOULD BE EQUAL TO 
THE SQUARE ROOT
OF 8 + U TO THE FIFTH.
BUT OUR GOAL HERE IS NOT TO FIND 
H PRIME OF U.
OUR GOAL IS TO FIND 
H PRIME OF T.
SO IF WE APPLY THE CHAIN RULE, H 
PRIME OF T, OR IF WE WANT, DHDT,
IT WOULD BE EQUAL TO DHDU, 
WHICH WE JUST FOUND, x DUDT.
WHICH MEANS H PRIME OF T 
IS EQUAL TO, AGAIN, DHDU,
WHICH WE HAVE HERE,
THE SQUARE ROOT OF 
8 + U TO THE FIFTH x DUDT.
WELL, U PRIME IS DUDT, 
SO x 4T TO THE THIRD.
BUT NOTICE HOW WE HAVE THIS 
IN TERMS OF U AND T, NOT JUST T.
SO NOW WE'LL MAKE A FINAL 
SUBSTITUTION FOR U TO THE FIFTH.
REMEMBER, U = T TO THE FOURTH.
SO H PRIME OF T WOULD BE EQUAL 
TO 4T TO THE THIRD
x THE SQUARE ROOT 
OF THE QUANTITY 8 +
AND, AGAIN, 
NOW FOR U TO THE FIFTH,
SINCE U IS EQUAL TO 
T TO THE FOURTH,
THIS WOULD BE T TO THE FOURTH 
TO THE FIFTH,
OR T TO THE TWENTIETH.
SO U TO THE FIFTH IS EQUAL TO 
U TO THE TWENTIETH,
SO WE HAVE THE SQUARE ROOT 
OF THE QUANTITY
8 + T TO THE TWENTIETH.
I HOPE YOU HAVE FOUND THIS 
HELPFUL.
