- WELCOME TO 2 EXAMPLES OF IMPROPER INTEGRALS
WHERE WE HAVE DISCONTINUOUS INTEGRANDS
ON THE INTERVAL OF INTEGRATION.
LET'S BEGIN WITH A QUICK REVIEW.
REMEMBER, THERE ARE 2 TYPES OF IMPROPER INTEGRALS,
THE FIRST TYPE IS WHEN THE INTERVAL OF INTEGRATION
IS AN INFINITE INTERVAL
OR WHEN WE HAVE AN UPPER OR LOWER LIMIT OF INTEGRATION
OF EITHER A NEGATIVE INFINITY OR A POSITIVE INFINITY
OR IN SOME CASES BOTH.
AND WE CAN SEE IN THIS CASE WE WRITE THESE INTEGRALS
AS THE LIMITS OF INTEGRALS AS SEEN HERE.
BUT FOR OUR TWO EXAMPLES
OUR IMPROPER INTEGRALS HAVE DISCONTINUOUS INTEGRANDS
ON THE INTERVAL OF INTEGRATION.
WHICH MEANS OUR FUNCTION F OF X WILL BE EITHER DISCONTINUOUS
AT "A" THE LOWER LIMIT OF INTEGRATION
OR AT B THE UPPER LIMIT OF INTEGRATION.
SO IF THE INTEGRAND IS DISCONTINUOUS AT "A"
THE LOWER LIMIT OF INTEGRATION
WE CAN WRITE THE INTEGRAL AS THE LIMIT
AS C APPROACHED "A" FROM THE RIGHT SIDE
OR POSITIVE SIDE OF INTEGRAL FROM C TO B.
AND IF F OF X IS DISCONTINUOUS AT B
THE UPPER LIMIT OF INTEGRATION
WE CAN WRITE THE INTEGRAL AS A LIMIT
AS C APPROACHES B FROM THE LEFT SIDE OR NEGATIVE SIDE
OF THE INTEGRAL FROM "A" TO C.
THE THIRD CASE IS WHEN OUR FUNCTION F OF X
IS DISCONTINUOUS AT C
WHICH IS INSIDE THE INTERVAL FROM "A" TO B
WHICH WE CAN WRITE AS 2 SEPARATE INTEGRALS USING LIMITS.
AND THESE TYPES OF IMPROPER INTEGRALS
CAN BE A LITTLE MORE DIFFICULT TO IDENTIFY.
SO GOING BACK TO OUR FIRST EXAMPLE
NOTICE HOW OUR FUNCTION HERE WOULD BE UNDEFINED
OR DISCONTINUOUS AT X = 0.
WE CAN VERIFY THIS BY LOOKING AT THE GRAPH AS WE SEE HERE.
SO WE'RE GONNA WRITE THIS INTEGRAL AS A LIMIT
AS C APPROACHES 0 FROM THE POSITIVE SIDE
OR RIGHT SIDE OF THE INTEGRAL FROM C TO 6.
SO WE'LL HAVE THE LIMIT AS C APPROACHES 0
FROM THE POSITIVE SIDE OR RIGHT SIDE
OF THE INTEGRAL FROM C TO 6.
I'M GONNA GO AHEAD AND WRITE THIS
AS X TO THE POWER OF NEGATIVE 0.8.
SO WE CAN APPLY THE POWER RULE OF INTEGRATION.
SO NOW WE'LL INTEGRATE THIS AND THEN FIND THE LIMIT.
SO WE'LL HAVE THE LIMIT AS C APPROACHES 0 FROM THE RIGHT.
APPLYING THAT POWER RULE OF INTEGRATION
WE WOULD HAVE X TO THE POWER OF -0.8 + 1.
THAT WOULD BE X TO THE POWER OF 0.2 DIVIDED BY 0.2
AND OUR LOWER LIMIT OF INTEGRATION IS C,
THE UPPER LIMIT OF INTEGRATION IS 6.
FROM HERE NOTICE HOW WE HAVE 0.2 IN THE DENOMINATOR
WHICH IS 2/10 OR 1/5.
SO WE HAVE X TO THE 0.2 DIVIDED BY 1/5
WHICH IS THE SAME AS MULTIPLYING BY THE RECIPROCAL OF FIVE.
SO WE CAN GO AHEAD AND WRITE THIS AS THE LIMIT
AS C APPROACHES 0 FROM THE RIGHT OF 5,
X THE POWER OF 0.2.
SO WE HAVE THE LIMIT AS C APPROACHES 0
FROM THE RIGHT OF 5 X 6.
RAISE THE POWER OF 0.2 - 5 X C, RAISE THE POWER OF 0.2.
NOW WE CAN DETERMINE THIS LIMIT.
NOTICE AS C APPROACHES 0
5 X C TO THE 0.2 IS GOING TO APPROACH 0
AND THIS IS NOT EFFECTED BY C
AND, THEREFORE, EQUAL TO 5 TIMES 6 RAISE THE POWER OF 0.2.
REMEMBER, THIS IS 2/10 OR 1/5.
SO WE COULD WRITE THIS AS 5 X 6 TO THE 1/5
OR IF WE WANTED TO 5 X THE 1/5 ROOT OF 6.
OR AS A DECIMAL APPROXIMATION
THIS WOULD BE APPROXIMATELY 7.1548.
AND NOW THAT WE'VE DETERMINED THIS DEFINITE INTEGRAL IS = TO
5 TIMES THE 1/5 ROOT OF 6 WE ALSO KNOW IT CONVERGES.
IF WE GO BACK TO THE GRAPH FOR A MOMENT
BECAUSE THE FUNCTION IS NON-NEGATIVE ON THIS INTERVAL
WE'VE DETERMINED THAT THIS AREA OF THIS UNBOUNDED REGION
ON THE INTERVAL FROM 0 TO 6
= 5 X THE 1/5 ROOT OF 6 SQUARE UNITS.
LET'S TAKE A LOOK AT A SECOND EXAMPLE.
THIS ONE LOOKS VERY SIMILAR.
AGAIN, NOTICE HOW THE FUNCTION IS DISCONTINUOUS OR UNDEFINED
AT X = 0.
AGAIN IF WE WANT TO WE CAN VERIFY THIS
BY LOOKING AT THE GRAPH.
NOTICE HOW WE HAVE INFINITE DISCONTINUITY AT X = 0.
SO ONCE AGAIN WE'LL WRITE THIS AS THE LIMIT
AS C APPROACHES 0 FROM THE RIGHT OR POSITIVE SIDE
OF INTEGRAL FROM C TO 3.
AGAIN LET'S WRITE THIS AS X TO THE POWER OF -1.2
AND NOW WE'LL INTEGRATE.
SO IF WE ADD ONE TO THE EXPONENT
WE WOULD HAVE X TO THE POWER OF -0.2 DIVIDED BY -0.2.
TO SIMPLIFY THIS FRACTION HERE WE HAVE DIVIDED -2/10
OR DIVIDED BY -1/5
WHICH IS THE SAME AS MULTIPLYING BY THE RECIPROCAL
AND MULTIPLYING BY -5.
SO WE CAN WRITE THIS AS THE LIMIT AS C APPROACHES 0
FROM THE RIGHT OF -5
AND THEN BECAUSE THE EXPONENT IS NEGATIVE
WE'LL MOVE THIS TO THE DENOMINATOR.
SO IT'D BE -5 DIVIDED BY X TO THE POWER OF +0.2.
SO NOW WE HAVE THE LIMIT
AS C APPROACHES 0 FROM THE RIGHT OF -5
DIVIDED BY 3 RAISED TO THE POWER OF 0.2
MINUS -5 DIVIDED BY C RAISED TO THE POWER OF 0.2.
NOW, WE'LL EVALUATE THE LIMIT.
WE'LL NOTICE HOW THIS FIRST TERM IS NOT AFFECTED BY C.
SO WE CAN WRITE THIS AS -5 DIVIDED BY,
THIS IS 3 TO THE 2/10 OR 3 TO THE 1/5 POWER
WHICH WOULD BE THE SAME AS 1/5 ROOT OF 3.
THEN WE'LL HAVE -,
NOW NOTICE HOW THIS FRACTION HERE AS C APPROACHES 0
OUR DENOMINATOR IS GETTING SMALLER AND SMALLER
AND WHILE OUR NUMERATOR IS FIXED AT -5.
WELL, NEGATIVE 5 DIVIDED BY
SMALLER AND SMALLER POSITIVE VALUES
WOULD APPROACH NEGATIVE INFINITY.
SO THIS APPROACH IS NEGATIVE INFINITY
AND THEREFORE WE HAVE - NEGATIVE INFINITY
AND THEREFORE THIS LIMIT IS APPROACHING POSITIVE INFINITY
WHICH DOES NOT EXIST
AND THEREFORE OUR -- INTEGRAL IS DIVERGENT.
SO THE DEFINITE INTEGRAL IS DIVERGENT
AND APPROACHES POSITIVE INFINITY.
WHICH MEANS IF WE GO BACK TO OUR GRAPH FOR A MOMENT
NOTICE THAT ONCE AGAIN THE FUNCTION IS NOW NEGATIVE
ON THE INTERVAL FROM 0 TO 3
BUT BECAUSE THIS -- INTEGRAL IS DIVERGENT
AND APPROACHES POSITIVE INFINITY
WE CANNOT DETERMINE THE AREA OF THIS UNBOUNDED SHADED REGION
THAT'S APPROACHING POSITIVE INFINITY.
I HOPE YOU FOUND THIS EXPLANATION HELPFUL.