I still don't understand why for the example for convergence we used the right endpoint of the intervals to determine the area of the squares, and for the example for divergence we used the left end point of the interval to determine the area of the block... it seems to arbitrary... if I had made squares using the right end points for the example with divergence, then the sum of the area of all the squares would be less than the area under the curve 1/x... can someone please clarify this?
+Alive4Metal If you choose the left end point, then the integral will be less than the area of the rectangles, and you can use this to prove that sum diverges by showing that the integral(which is smaller) diverges(meaning the integral is infinite. If you think that the series converges, than you use the right end point, and the integral will be greater the sum of the series. If the integral is not infinite, than the sum of the series must also converge,since it is smaller than the integral. Why does this work? This is a test for positive infinite series, so you can factor out the negative if they're negative numbers. Remember, the numbers being added must end up being zero eventually(even at infinity), therefore the rectangles must be getting smaller and smaller.
When you're applying the Integral Test, you really don't need to worry about that, because if you add or subtract a finite number of terms in any series, that won't influence whether the series converges or diverges.
This is where mathematics shows you the future. In other words, from here on out, logic, theorems, and justification.
Why is there no practice problems for integral test and the root test on khan academy?
Thanks from Turkey Mr!
Clearly put and a very good lesson.
What if i decide to draw the rectangle leftwards for 1/x will that be now not conclusive that 1/x is convergent or divergent
I still don't understand why for the example for convergence we used the right endpoint of the intervals to determine the area of the squares, and for the example for divergence we used the left end point of the interval to determine the area of the block... it seems to arbitrary... if I had made squares using the right end points for the example with divergence, then the sum of the area of all the squares would be less than the area under the curve 1/x... can someone please clarify this?
+Alive4Metal If you choose the left end point, then the integral will be less than the area of the rectangles, and you can use this to prove that sum diverges by showing that the integral(which is smaller) diverges(meaning the integral is infinite.
If you think that the series converges, than you use the right end point, and the integral will be greater the sum of the series. If the integral is not infinite, than the sum of the series must also converge,since it is smaller than the integral.
Why does this work? This is a test for positive infinite series, so you can factor out the negative if they're negative numbers. Remember, the numbers being added must end up being zero eventually(even at infinity), therefore the rectangles must be getting smaller and smaller.
When you're applying the Integral Test, you really don't need to worry about that, because if you add or subtract a finite number of terms in any series, that won't influence whether the series converges or diverges.
What are Integral test applications of iur daily Life...anyone?
You're a God
good
But isn't 1/x convergent? How could it = infinity
1/x is convergent but the integral of 1/x which is ln(x) is not convergent.
It approaches infinity too slowly.
No graphical explanation?
Useless.
Did you even watch the whole thing
The graphical part is provided at 6:40