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# Find the limit?

Find the limit of ∫ sqrt(x^5+8)dx from 1 to 1+h all over h as h approaches to 0.

### 2 Answers by Expert Tutors

Murtaza N. | Math, Physics, Computer Programming, & Test Prep!Math, Physics, Computer Programming, & T...
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Hi Sun,

There's a way to see how to do this, but I'm gonna make the problem a bit more abstract first...  Let's say the problem was

"Find the limit of ∫ f(x) dx from a to a+h all over h as h approaches 0." (a is a number, in your case it's 1)

Then, ∫ f(x) dx = F(x) + C (where F(x) is an antiderivative of f(x), and the integral of f(x) from x=a to x=a+h  is given by F(a+h) - F(a).  (1st Fundamental Theorem of Calculus)

Thus, since you have to divide this integral by h, you have to find the limit of:
lim h→0 (F(a+h) - F(a)) / h .

Here's where it gets cool.  You probably recognize that that limit is equal to F'(a), a.k.a. f(a).

In the case of your problem, a=1 so all you have to do is plug in x = 1 into sqrt(x^5+8)!  That's where the 3 came from haha.

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I also wanna note that Steve's idea with the rectangles is good:  the integral value *does* approach 0.  As the rectangle closes in on x = 1, its height is 3 since that's the y-value you get when you plug in x = 1 into sqrt(x^5+8).  The width of the rectangle is h.  So the integral's value tends to 3h.  But the problem asks you to divide by h, which is why the limit approaches 3.

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I feel like my presentation was kinda confusing so please down-thumb me if you agree!  Thx :)

Your explanation has a sparkling clarity!

Thanks!
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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Haven't worked in integral calculus for awhile, but here are my thoughts.

The integral is the sum of rectangles whose width are dx and height are f(x) = sqrt(x^5+8).

With the bounds h apart and h approaching zero, wouldn't both bounds approach the same value of 1, and then the base widths of all the rectangles approach zero? So the integral value approaches zero?