- WE WANT TO SKETCH THE REGION ENCLOSED BY Y EQUALS 2X
AND Y EQUALS 3X SQUARED,
THEN FIND THE AREA OF THE REGION.
IF WE GRAPH THESE 2 FUNCTIONS ON THE COORDINATE PLANE,
THIS WOULD BE OUR GRAPH
AND WE'RE LOOKING FOR THE AREA OF THIS REGION
SHADED HERE IN BLUE
WHICH IS THE REGION ENCLOSED BY THE 2 FUNCTIONS.
WHEN DETERMINING THE AREA BETWEEN 2 FUNCTIONS,
WE WANT TO FIND THE INTEGRAL OF F OF X - G OF X FROM "A" TO B
AND WHERE F OF X WOULD BE THE TOP FUNCTION
AND G OF X WOULD BE THE BOTTOM FUNCTION,
AND "A" AND B WOULD BE THE X COORDINATES
OF THE POINTS OF INTERSECTION.
SO WHAT'S HAPPENING HERE
IS WE'RE FINDING THE AREA BELOW THE TOP FUNCTION
AND ABOVE THE X-AXIS,
AND IT'S ATTRACTING THE AREA BELOW THE LOWER FUNCTION
AND ABOVE THE X-AXIS
LEAVING US WITH THE AREA BETWEEN THE 2 FUNCTIONS.
SO GOING BACK TO OUR EXAMPLE,
NOTICE HOW THE LINEAR FUNCTION IS ON TOP
AND THE QUADRATIC FUNCTION IS ON THE BOTTOM,
SO WE'LL HAVE THE INTEGRAL OF 2X - 3X SQUARED.
BUT NOW, WE ALSO HAVE TO FIND THE LIMITS OF INTEGRATION
WHICH WILL BE THE X COORDINATES
OF THESE 2 POINTS OF INTERSECTION.
TO FIND THESE X VALUES,
WE'LL SOLVE THIS AS A SYSTEM OF EQUATIONS
AND WE'LL SOLVE THIS USING SUBSTITUTION,
WHERE IF Y EQUALS 2X AND Y EQUALS 3X SQUARED,
THEN 2X MUST EQUAL 3X SQUARED.
SO TO SOLVE THIS,
WE'LL SET IT EQUAL TO 0 BY SUBTRACTING 2X ON BOTH SIDES,
THAT WOULD GIVE US 0 = 3X SQUARED - 2X.
WE CAN SOLVE THIS BY FACTORING THE COMMON FACTOR HERE OF X
LEAVING US WITH THE QUANTITY 3X - 2.
SO EITHER X = 0 OR 3X - 2 = 0.
SO WE HAVE X = 0 AND HERE WE WOULD ADD 2 AND DIVIDE BY 3,
THAT WOULD BE X = 2/3
WHICH MEANS, "A," THE LOWER LIMIT OF INTEGRATION WOULD BE 0
AND, B, THE UPPER LIMIT OF INTEGRATION WOULD BE 2/3.
NOW, LET'S SET UP THE DEFINITE INTEGRAL TO FIND THIS AREA.
WE'LL HAVE THE INTEGRAL OF,
AGAIN, THE TOP FUNCTION - THE BOTTOM FUNCTION
WHICH WAS THE LINEAR FUNCTION - THE QUADRATIC FUNCTION
INTEGRATED FROM 0 TO 2/3.
SO THE ANTIDERIVATIVE OF 2X WOULD BE 2 x X TO THE 2nd
DIVIDED BY 2 - THE ANTIDERIVATIVE OF 3X SQUARED,
THAT'D BE 3 x X TO THE 3rd DIVIDED BY 3.
WE EVALUATE THIS AT 2/3 AND 0 AND THEN FIND THE DIFFERENCE,
BUT LET'S SIMPLIFY FIRST.
THIS WOULD BE X SQUARED - X CUBED.
SO WHEN X IS 2/3, WE'D HAVE 2/3 SQUARED - 2/3 CUBED -,
WHEN X IS 0, WE JUST HAVE 0 SQUARED - 0 CUBED,
SO HERE WE'RE GOING TO HAVE, WHAT, 4/9 -, THIS WOULD BE 8/27.
OF COURSE, THIS WOULD JUST BE 0,
OBTAINING OUR COMMON DENOMINATOR WHICH WOULD BE 27,
MULTIPLY 4/9 BY 3/3.
SO WE HAVE 12/27 - 8/27 WHICH WOULD GIVE US 4/27
WHICH WOULD BE THE AREA OF THE REGION ENCLOSED BY Y EQUALS 2X
AND Y EQUALS 3X SQUARED
WHICH IS THIS AREA HERE.
I HOPE YOU FOUND THIS HELPFUL.