WHY DO WE NEED INTEGRATION?

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Presentation transcript:

WHY DO WE NEED INTEGRATION?

INTEGRATION 1.FIGURE OUT FUNCTION FROM ITS RATE OF CHANGE 2. ADD VALUES OF A FUNCTION, ESPECIALLY INFINITELY MANY

SLOPE FIELD SHOWING RATE OF CHANGE Each little arrow represents direction and size (usually by its length) of slope of a function at that point This is actually slope field of a bunch of curves, each the graph of a function

Examples Velocity at a point P’(t) = Growth of population in one year (very approximately) Knowing P’(t) for several years in a row, could try to figure out P(t) Similarly for revenue function, etc Velocity at a point V(t) = S’(t) where S(t) is position at time t Could try to figure out S(t) using V(t)

INTEGRAL OF Y = X^2 LET US SAY THE SLOPE OF A CURVE F(X) IS GIVEN BY X^2 FOR ALL VALUES OF A FUNCTION SO F’(X) = X^2 THEN JUST BY GUESSING WE SEE THAT F(X) = X^3/3. (DERIVATIVE OF X^3/3 = 3X^2/3 = X^2)

ADDING VALUES OF A FUNCTION

PARABOLA Y = X^2 THIS SHOWS GRAPH FOR X =0 TO X=1 INFINITE NUMBER OF X-VALUES FROM 0 TO 1 WANT TO ADD Y-VALUES (OR FIND AREA UNDER CURVE)

IDEA: CHOP AREA INTO BLOCKS AREA UNDER CURVE APPROXIMATED BY AREA OF BLOCKS AREA OF UPPER BLOCKS = (.5)(.25) +(.5)1 = .625 AREA OF LOWER BLOCKS = .25(.5) = .125 ACTUAL AREA = .33

CHOPPING INTO MORE BLOCKS NOW WE CHOP AREA INTO FOUR BLOCKS, EACH OF LENGTH .25 TOTAL AREA OF UPPER (.25)(.0625+.25+.5625+1) =0.46875 = 0.47 APPROX TOTAL AREA OF LOWER (.25)(.0625+.25+.5625) =0.21875 = 0.22 APPROX GETTING CLOSER!

CHOPPING EVEN FINER AREA OF UPPER BLOCKS (EACH OF LENGTH 0.125) = 0.40 APPROX AREA OF LOWER BLOCKS = 0.27 APPROX ACTUAL AREA = 0.33 -- EVEN CLOSER!

AND EVEN FINER…! AREA OF UPPER = 0.37 APPROX AREA OF LOWER ACTUAL AREA = 0.33 -- ALMOST THERE!

MAKING IT EXACT AS YOU CHOP INTO MORE AND MORE BLOCKS THE AREA OF BLOCKS GETS CLOSER AND CLOSER TO ACTUAL AREA

MAKING IT EXACT HOW DO YOU MAKE IT EXACT?

ANSWER: TAKE LIMITS! AREA = LIMIT OF SUM OF AREA OF BLOCKS, AS N GOES TO INFINITY N BEING NUMBER OF BLOCKS

NOW FOR THE PUNCHLINE WHAT IS THE RELATION BETWEEN AREA UNDER Y = X^2 OBTAINED USING BLOCKS AND INTEGRAL OF X^2 = X^3/3 (BECAUSE (X^3/3)’ = X^2)

GRAND CONCLUSION (END OF MOVIE) X=0 TO X=1 WE HAVE (1^3/3)-(0^3/3) = 1/3 = 0.33 (APPROX) BY USING BLOCKS WE GOT AREA = 0.33 (APPROX) SO THEY ARE THE SAME! THIS IS CALLED FUNDAMENTAL THEOREM OF CALCULUS