- WELCOME TO A LESSON ON INTEGRATING FUNCTIONS
OF TWO VARIABLES.
WE CAN INTEGRATE FUNCTIONS OF TWO VARIABLES
IN A SIMILAR WAY WE DETERMINE PARTIAL DERIVATIVES
OF A FUNCTION OF TWO VARIABLES.
MEANING IF INTEGRATING WITH RESPECTS TO X
WE'LL TREAT Y AS A CONSTANT AND IF INTEGRATING
WITH RESPECTS TO Y WE'LL TREAT X AS A CONSTANT.
LET'S SEE WHAT HAPPENS WHEN WE TAKE A FUNCTION,
FIND IT'S FIRST ORDER OF PARTIALS
AND THEN INTEGRATE WITH RESPECTS TO X
AS WELL AS WITH RESPECTS TO Y.
SO IF WE'RE GIVEN THE FOLLOWING FUNCTION IN TWO VARIABLES
THE PARTIAL DERIVATIVE OF F WITH RESPECTS X WOULD BE 9X SQUARED,
THIS WOULD BE 0, THIS WOULD BE - 1,
THIS WOULD BE Y, AND THIS WOULD BE 0
SO HERE'S THE PARTIAL OF F WITH RESPECT TO X.
NOW LETS' GO AHEAD AND SEE IF WE RECOVER THE FUNCTION
IF WE INTEGRATE THIS PARTIAL DERIVATIVE WITH RESPECTS TO X.
SO WE'D HAVE 9 x X TO THE 3rd/3 - X + AGAIN TREATING Y
AS A CONSTANT THIS WOULD GIVE US Y x X.
LET'S GO AHEAD AND SIMPLIFY THIS.
WE'RE GOING TO HAVE 3X CUBED - X + XY.
NOW IF WE COMPARE THIS ANTIDERIVATIVE
WITH THE ORIGINAL FUNCTIONS IN TWO VARIABLES
NOTICE WE DON'T RECOVER THE ENTIRE FUNCTION.
WE'RE MISSING THIS TERM HERE AS WELL AS THIS TERM HERE.
SO INSTEAD OF JUST PUTTING + C LIKE WE NORMALLY DO
WHEN INTEGRATING FUNCTIONS OF 1 VARIABLE
WE'RE GOING TO PUT + C OF Y TO MAKE SURE
THAT WE RECOGNIZE THAT WE MAY NOT HAVE RECOVERED
THE ENTIRE FUNCTION IN TERMS OF Y.
NOW LET'S DETERMINE THE PARTIAL OF F WITH RESPECT TO Y.
HERE WE'D HAVE 0, HERE WE'D HAVE 8/Y CUBED.
THIS WOULD BE 0.
THIS WOULD BE + X AND THIS WOULD BE 2.
REMEMBER WE'RE TREATING X AS A CONSTANT
AND NOW WE'RE GOING TO INTEGRATE THIS WITH RESPECTS TO Y,
AGAIN TREATING X AS A CONSTANT.
SO HERE WE'D HAVE 8 x Y TO THE - 2 DIVIDED BY -2
+ AGAIN, X IS A CONSTANT SO WE'D HAVE X x Y
AND THIS WOULD BE + 2Y.
LET'S GO AHEAD AND SIMPLIFY THIS.
THIS WOULD BE -4/Y SQUARED + XY + 2Y.
COMPARING THIS FUNCTION TO THE ORIGINAL FUNCTION
NOTICE AGAIN WE'RE MISSING SOME OF THE X TERMS
SO OUR CONSTANT OF INTEGRATION WILL BE C OF X.
LET'S TAKE A LOOK AT ANOTHER EXAMPLE.
HERE WE HAVE A DEFINITE INTEGRAL OF A FUNCTION OF TWO VARIABLES
AND WHAT I WANT YOU TO NOTICE HERE
IS THAT WE'RE INTEGRATING WITH RESPECTS TO X
BUT THE LIMITS OF INTEGRATION ARE IN TERMS OF Y.
THIS WILL BE IMPORTANT LATER ON.
SO BECAUSE WE'RE INTEGRATING WITH RESPECTS TO X
WE'RE GOING TO TREAT Y AS A CONSTANT.
SO HERE WE'RE GOING TO HAVE 2 x X SQUARED/2
AND THEN TREATING Y AS A CONSTANT WE'LL HAVE
- 3Y x X AND THEN WE HAVE OUR LIMITS OF INTEGRATION.
LET'S GO AHEAD AND SIMPLIFY THIS.
WE'RE GOING TO HAVE X SQUARED - 3XY
AND BECAUSE WE INTEGRATED WITH RESPECTS TO X
WE'RE GOING TO REPLACE X WITH 2Y AND 0.
SO WE'LL HAVE 2Y SQUARED - 3 x 2Y x Y
AND OVER HERE X IS GOING TO BE ZERO.
SO WE'LL HAVE ZERO HERE AND ZERO HERE.
SO WE'RE GOING TO HAVE 4Y TO THE 2nd.
THIS IS GOING TO BE - 6Y SQUARED SO THE RESULT IS -2Y SQUARED.
LET'S TAKE A LOOK AT ANOTHER EXAMPLE.
HERE WE'RE INTEGRATING WITH RESPECTS TO Y
SO WE'LL TREAT X AS A CONSTANT,
AND AGAIN NOTICE OUR LIMITS OF INTEGRATION
ARE IN TERMS OF X WHEN WE INTEGRATE WITH RESPECTS TO Y.
SO IF X IS A CONSTANT
WE'RE TRYING TO FIND THE ANTIDERIVATIVE OF 1/Y
THAT'S GOING TO BE NATURAL LOG Y.
SO OUR ANTIDERIVATIVE IS X SQUARED NATURAL LOG Y
AND WE'LL EVALUATE THIS AT 2X AND 1
AND AGAIN, BECAUSE WE INTEGRATED WITH RESPECTS TO Y
WE'RE REPLACING Y WITH 2X AND Y WITH 1.
SO WE'RE GOING TO HAVE X SQUARED NATURAL LOG 2X - X SQUARED
NATURAL LOG 1.
WELL NATURAL LOG 1 = 0 SO WE'RE LEFT WITH X SQUARED
NATURAL LOG 2X AND NOTICE WE HAVE A FUNCTION
IN TERMS OF X NOW.
NOW THAT WE'VE INTEGRATED FUNCTIONS IN TWO VARIABLES
I WANT TO GIVE YOU A PREVIEW OF WHERE WE'RE HEADED.
NOW THAT WE CAN INTEGRATE A FUNCTION OF TWO VARIABLES,
WE'LL BEGIN A STUDY OF THE INTEGRAL OF AN INTEGRAL
OR A DOUBLE INTEGRAL ALSO CALLED A ITERATED INTEGRAL.
SO IF WE TAKE A INTEGRAL OF AN INTEGRAL
WHAT WE'RE GOING TO DO IS EVALUATE
THE INNER INTEGRAL FIRST AND THEN EVALUATE
THE OUTER INTEGRAL.
SO AS YOU CAN SEE FROM THIS NOTATION BELOW
THIS IS WHAT A DOUBLE INTEGRAL LOOKS LIKE.
SO IF IN THIS FORM WHAT WE'RE GOING TO DO
IS INTEGRATE WITH RESPECTS TO Y FIRST
AND THEN INTEGRATE WITH RESPECTS TO X,
AND IF IT'S IN THIS FORM
WE'LL INTEGRATE WITH RESPECTS TO X FIRST
AND THEN WITH RESPECTS TO Y. SO WE'LL WORK OUR WAY INSIDE OUT
AND I'D LIKE TO FINISH BY TALKING ABOUT THE CONNECTION
BETWEEN INTEGRATING A FUNCTION OF ONE VARIABLE
AND A DOUBLE INTEGRAL OF A FUNCTION OF TWO VARIABLES.
WE KNOW FROM CALC 1,
IF WE INTEGRATE A CONTINUOUS NONNEGATIVE FUNCTION
ON THE INTERVAL FROM "A" TO B AS WE SEE HERE
IT WILL GIVE US THE AREA UNDER THE CURVE
BOUNDED BY THE X AXIS ON THE GIVEN INTERVAL.
WELL NOW IF WE EXTEND THIS TO A FUNCTION OF TWO VARIABLES
THAT IS CONTINUOUS AND NONNEGATIVE
OVER A REGION IN THE XY PLANE
IT'LL GIVE US THE VOLUME BOUNDED BY THE XY PLANE
AND THE SURFACE DEFINED BY F OF XY.
NOW THIS IS A LOT OF NEW NOTATION.
THIS IS TELLING US WE'RE GOING TO FIND THE DOUBLE INTEGRAL
OVER A SPECIFIED REGION IN THE XY PLANE
OF A FUNCTION OF TWO VARIABLES.
NOTICE HERE IT STATES WE'RE INTEGRATING
WITH RESPECTS TO "A."
WELL THIS REALLY REPRESENTS EITHER DX/DY OR DY/DX
BASED UPON WHETHER WE WILL INTEGRATE
WITH RESPECTS TO X FIRST OR WITH RESPECTS TO Y FIRST.
SO WE'LL TALK A LOT MORE ABOUT THIS IN THE NEXT SEVERAL VIDEOS
BUT I WANTED TO MENTION THE CONNECTION
BETWEEN INTEGRATING A FUNCTION OF ONE VARIABLE
AND DETERMINING A DOUBLE INTEGRAL OF A FUNCTION
OF TWO VARIABLES. THANK YOU FOR WATCHING.