Symmetric Antiderivative?
Find h → 0 lim h 1 ∫ 4 − h 4 + h x 3 − 2 x + 5 d x
The answer is 122.
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1 solution
The function inside integral would be 6 1 on average around the interval 4 − h to 4 + h (which is infinitely small). Hence, integral can be approximated as 6 1 × 2 h and answer is 122.
Note to self: subtraction of a negative is addition.
There is a proof that lets you skip Binomial Theorem bashing. Can you figure it out?
Hint: What if the problem asked for h → 0 lim 2 h 1 ∫ 4 − h 4 + h x 3 − 2 x + 5 d x ?
You don't need to evaluate the integral. The other approach is to use L'Hopital rule (not sure about the spelling) and differentiation under the integral sign.
EDIT: Woops, I did not read Trevor's comment, maybe that's what he is talking about.
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Yes, the two intended solutions for this problem were l'Hospital's rule and the method I gave the hint for. Can you figure it out with the hint?
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Just substitute x = 4 into ( x 3 − 2 x + 5 ) then times 2 . This is because you're finding the average of the function f ( x ) = x 3 − 2 x + 5 from 4 − h to 4 + h .
Multipliying and dividing the limit proposed by 2, 1/2h times the above integral is the definition of symmetric derivative (in the interval (4-h,4+h), instead of (4,4+h)) of the integral of f(t)=x^3-2x+5. So, the limit in question is simply 2xlim_{h->0} [(4+h)^3-2(4+h)+5] = 122 , or likewise, 2xlim_{h->0} [(4-h)^3-2(4-h)+5] = 122 . I posted this comment to point out that is not neccesary (or even impossible) to know the antiderivative of the subintegral function... =0)
First, let us evaluate the integral. We have that ∫ x 3 − 2 x + 5 d x = 4 1 x 4 − x 2 + 5 x .
Now we must evaluate this at x = 4 + h and x = 4 − h , then subtract the latter from the former. Doing so with some messy Binomial Theorem bash, we obtain the nice result 4 1 x 4 − x 2 + 5 x ∣ ∣ ∣ ∣ 4 − h 4 + h = 8 h 3 + 1 2 2 h
We want to find h → 0 + lim h 8 h 3 + 1 2 2 h . The expression simplifies to 8 h 2 + 1 2 2 ; therefore, our limit is simply 1 2 2 .
An interesting note: the requirement of approaching from the positive side is not needed. Either side approaches 1 2 2 .