Question

Use the comparison theorem to determine whether the integral of sec^2x / xsqrt(x) dx from 0 to 1 is convergent or divergent

Answer 1

Hint : since \(|\sec x| \ge 1\), we have :

\(\dfrac{\sec^2x}{x\sqrt{x}} \ge\dfrac{1}{x\sqrt{x}} \)

Answer 2

I understood that, but how do I prove it, and I want to understand how you can determine why something diverges or is convergent.

Answer 3

good question,
let me ask you a quick question, look at below sum and tell me if you think it converges :
\[ 1+1+1+1+\cdots \]

Answer 4

yes because it is greater than one

Answer 5

so are you saying that sum converges ?

Answer 6

that's what I thought, am I wrong?

Answer 7

yes, loosely speaking, "converging sum" means that the sum approaches some value as you increase the number of terms that you add

Answer 8

\[ 1+1+1+1+\cdots \]

look at above sum again, does it seem like approaching any finite number ?

Answer 9

maybe take the partial sums and convince yourself that the sum is NOT approaching any fixed number :

1+1 = ?
1+1+1 = ?
1+1+1+1 = ?
1+1+1+1+1 = ?

Answer 10

I get it now, the sum will be approaching infinity and so therefore it has to be diverging

Answer 11

Exactly! You got it mostly! but thats one way to look at it.

Answer 12

look at below sum and tell me if it converges or diverges :

\[1-1+1-1+1-1+\cdots \]

Answer 13

i would think it still diverges because it is not stable

Answer 14

Yep! now you know what the words, "converge" or "diverge" mean exactly.

Lets try one more last example

Answer 15

Look at below sum, does it converge ?
\[32+ 16+ 8+4+2+1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots\]

Answer 16

no because regardless of the value it will still be constantly increasing and so it is not a standard limit

Answer 17

think again, here is a hint :
geometric series

Answer 18

I don't understand, I do notice that it is all going down at a ratio of 1/2

Answer 19

Yes, remember the partial sum formula for geometric series ?

Answer 20

if not, don't bother... we don't really need to know about geometric series as of now...

Answer 21

I dont think I was taught that

Answer 22

that makes sense, okay we're done with examples. lets get back to the original problem

Answer 23

so, with integerals also, we define convergence and divergence similalrly :

if the integral evaluates to a fixed number, we say the integral "converges"

otherwise, we say the integral diverges

Answer 24

I did the steps and it ended up being -2, but I know that it is supposed to diverge

Answer 25

what has ended up being -2 ?

Answer 26

I'll write the steps that I took

Answer 27

Okay

Answer 28

\[\lim_{t \rightarrow 0} \int\limits_{t}^{1} 1/(x \sqrt{x}) \]
\[\int\limits_{t}^{1} x^-3/2\]
\[\lim_{t \rightarrow 0} \left( -2x^(-\frac{ 1 }{ 2 } )\right)\]
\[\lim_{t \rightarrow 0} \left[ -2 - (-2t^(-1/2) \right]\]
= -2 when t goes to 0

Answer 29

Looks very good till the last line
there is a huge mistake in the last line when you evaluate the limit

Answer 30

\(\lim\limits_{t \rightarrow 0} \left[ -2 - (-2t^{-1/2} \right]\)

is same as

\(\lim\limits_{t \rightarrow 0} \left[ -2 + \dfrac{2}{\sqrt{t}} \right]\)

yes ?

Answer 31

i thought because it went to 0 it got eliminated

Answer 32

Now if u evaluate the limit, you would see that you get infinity
so the integral doesn't converge.

Answer 33

look at below limit :

\(\lim\limits_{t\to 0} \dfrac{1}{t}\)

what does it evaluate to ?

Answer 34

1/0.1 = ?

1/0.01 = ?

1/0.00001 = ?

Answer 35

infinity >< I messed up, I was thinking like 0/1 = 0

Answer 36

Happens... Okay, so now we know that the integral \(\int\limits_0^1 \dfrac{1}{x\sqrt{x}}\,dx\) doesn't converge.

Answer 37

what does that tell us about the integral \(\int\limits_0^1 \dfrac{\sec^2x}{x\sqrt{x}}\,dx\) ?

Answer 38

that it diverges also based on f(x) > g(x) and there is the concept that if g(x) diverges then so does f(x)

Answer 39

Looks good!
when free just take a quick look
http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx

Answer 40

Thank you so much for your help, my calculus 2 exam is like in 8 hours. I really appreciate it.

Answer 41

np, You seem to be mostly ready with the material ! good luck !