Simple Improper Integral Question (just a question of concept understanding)

Click For Summary
The integral ∫_0^1 sin(x)/x dx is being analyzed for convergence. It is noted that the lower limit of zero presents a removable discontinuity. The limit as x approaches zero of sin(x)/x equals 1, indicating that the function is bounded near this point. This suggests that the integral is convergent due to the function being integrable over the interval. The conclusion drawn is that the integral converges.
neden
Messages
18
Reaction score
0

Homework Statement



I am to determine whether the following integral is convergent or divergent

<br /> \int_0^1 \frac{sin(x)}{x}<br />

From what I hear since, lower limit is zero there is a removable discontinuity.
Thus just because of this, it is convergent? Can someone let me know if this
is correct.
 
Last edited:
Physics news on Phys.org
Do you know that:

lim(x-&gt;0) \frac{sin(x)}{x}=1

?

If yes, then you can see that the function is bounded on the interval and therefore integrable.
 
Oh thanks!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
11K
Replies
2
Views
1K
Replies
7
Views
5K