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

Answer: 3

Please show all your work. Thanks.

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

Answer: 3

Please show all your work. Thanks.

Tutors, please sign in to answer this question.

Hialeah, FL

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)

"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.

***

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.

***

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

Westford, MA

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?

wups ... missed the "over h" as a denominator!

Mark D.

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