- WELCOME TO AN EXAMPLE OF AN IMPROPER INTEGRAL.
WE WANT TO DETERMINE IF THE INTEGRAL CONVERGES OR DIVERGES.
IF IT CONVERGES, WE WANT TO EVALUATE THE INTEGRAL.
IF IT DIVERGES, WE WANT TO INDICATE IF IT APPROACHES
EITHER POSITIVE OR NEGATIVE INFINITY.
SO IF WE TAKE A LOOK AT OUR INTEGRAL
NOTICE HOW THE LOWER LIMIT OF INTEGRATION
IS NEGATIVE INFINITY.
SO THE INTERVAL OF INTEGRATION IS AN INFINITE INTERVAL
THIS IS THE REASON WHY WE HAVE AN IMPROPER INTEGRAL.
SO WE'RE GOING TO WRITE THIS AS THE LIMIT
AS LETS SAY "A" APPROACHES NEGATIVE INFINITY
SO WE'RE GOING TO SUBSTITUTE "A" FOR NEGATIVE INFINITY.
SO WE'LL HAVE THIS LIMIT OF THE INTEGRAL.
LET'S GO AHEAD AND FACTOR OUT THIS 3.
SO THE INTEGRAL WILL BE FROM "A" TO 2
AND THE INTEGRAND WOULD NOW
JUST BE 1 DIVIDED BY THE QUANTITY X SQUARED + 1.
SO NOW WE'LL INTEGRATE AND THEN DETERMINE THE LIMIT.
TO INTEGRATE WE'LL BE APPLYING THIS INTEGRATION FORMULA HERE.
NOTICE HOW WE CAN CHANGE THE ORDER OF THIS SUM
AND THEREFORE "A" SQUARED IS 1,
WHICH MEANS "A" IS 1 AND U SQUARED WOULD BE X SQUARED
SO U IS JUST X.
SO IF "A" IS 1 AND U IS X,
OUR ANTIDERIVATIVE WOULD BE JUST ARCTANGENT X + C
BUT BECAUSE WE HAVE A DEFINITE INTEGRAL WE'LL HAVE THE LIMIT
AS "A" APPROACHES NEGATIVE INFINITY OF 3 ARCTANGENT X,
AND WE STILL HAVE OUR LIMITS OF INTEGRATION FROM "A" TO 2.
SO NOW WE'LL HAVE THE LIMIT AS "A" APPROACHES NEGATIVE INFINITY
OF 3 ARCTANGENT 2 - 3 ARCTANGENT "A"
AND NOW TO EVALUATE THIS LIMIT
NOTICE HOW THIS FIRST TERM IS NOT AFFECTED BY "A"
SO THE LIMIT IS GOING TO BE EQUAL TO 3 ARCTANGENT 2
AND THEN - 3 ARCTANGENT "A"
AS "A" APPROACHES NEGATIVE INFINITY.
TO EVALUATE ARCTANGENT "A"
AS "A" APPROACHES NEGATIVE INFINITY,
WE'RE TRYING TO FIND AN ANGLE
THAT HAS A TANGENT FUNCTION VALUE
THAT WOULD BE APPROACHING NEGATIVE INFINITY.
SO TO DO THIS LETS ANALYZE THE GRAPH OF THE TANGENT FUNCTION.
SO AGAIN HERE WE HAVE Y = TANGENT X
AND NOTICE AS THE TANGENT FUNCTION VALUES
APPROACH NEGATIVE INFINITY IN THIS DIRECTION,
NOTICE HOW THE ANGLE IS APPROACHING -PI/2
AND THEREFORE WE'D HAVE - 3 x -PI/2.
SO THIS LIMIT DOES EXIST
AND THE EXACT VALUE WOULD BE 3 ARCTANGENT 2 + 3PI/2.
LET'S ALSO GET A DECIMAL APPROXIMATION THOUGH.
SO WE'LL GO TO THE GRAPHING CALCULATOR.
LET'S MAKE SURE THAT WE'RE IN RADIAN MODE
SO WE'LL PRESS THE MODE KEY.
NOTICE HOW RADIAN IS ALREADY HIGHLIGHTED.
SCROLL BACK TO THE HOME SCREEN
AND WE'LL TYPE IN 3 2nd TANGENT FOR ARCTANGENT 2 + 3PI/2.
PI KEY IS 2nd AND THEN THE EXPONENT KEY
OR THE CARAT KEY DIVIDED BY 2.
SO TO 4 DECIMAL PLACES THIS WOULD BE APPROXIMATELY 8.0338.
SO WE CAN SAY THIS INTEGRAL CONVERGES
AND THE INTEGRAL = 3 ARCTANGENT 2 + 3PI/2 OR THIS APPROXIMATION.
IF WE TAKE A LOOK AT THE GRAPH OF THE FUNCTION 3
DIVIDED BY THE QUANTITY X SQUARED + 1,
IT IS NONNEGATIVE ON THIS INTERVAL
AND THEREFORE THIS WOULD BE THE AREA UNDER THE CURVE
BOUNDED BY THE X AXIS.
SO HERE'S THE GRAPH OF OUR FUNCTION.
ON THE INTERVAL FROM NEGATIVE INFINITY TO 2
WE WOULD HAVE THIS SHADED REGION BOUNDED BY THE FUNCTION
IN THE X AXIS
AND THEREFORE THIS AREA = 3 ARCTANGENT 2
+ 3PI/2 SQUARE UNITS
WHICH IS APPROXIMATELY 8.0338 SQUARE UNITS.
I HOPE YOU FOUND THIS EXPLANATION HELPFUL.