integration core 3

Find the value of each of the following infinite integrals :
integrate 1/(4x+1)^2 dx with the limits infinity and 0.
i have integrated this to get :

(4x+1)^-1 / -4
Then,substituting in zero and zero,
i get :
-1/4 - - 1/4 = 0.
but the answer in the back of the book says : 1/4??

Reply 1

JemmaSimmons

Original post by elvss567

Integrating gives -1/4*(4x+1)-1 (Agree with you.)

Therefore the value of the infinite integral is:
-1/4[1/infinity -1/1)=-1/4[0+-1]= -1/4*-1 = 1/4

Hope this helps. :smile:

Edit: I took -1/4 out as a factor. Doing it without factorising gives (-1/4*1/infinity) - (-1/4*1)=0--1/4=1/4

(edited 6 years ago)

Reply 2

elvss567

Original post by JemmaSimmons

Integrating gives -1/4*(4x+1)-1 (Agree with you.)

Therefore the value of the infinite integral is:
-1/4[1/infinity -1/1)=-1/4[0+-1]= -1/4*-1 = 1/4

Hope this helps. :smile:

Edit: I took -1/4 out as a factor. Doing it without factorising gives (-1/4*1/infinity) - (-1/4*1)=0--1/4=1/4

how did you get zero for the infinity bit? what did you substitute in for x? is there an infinity button on the calculator or something? :smile:

Reply 3

JemmaSimmons

Original post by elvss567

how did you get zero for the infinity bit? what did you substitute in for x? is there an infinity button on the calculator or something? :smile:

1/2=0.5
1/1000=0.001
1/2000000=0.0000005

Pattern: As you divide 1 by a larger number the result gets smaller. This means that the value of 1/x tends towards 0 as x approaches infinity. So we replace 1/infinity with 0.

Think of the graph of 1/x.

Hope this helps. :smile:

(edited 6 years ago)

Reply 4

elvss567

Original post by JemmaSimmons

1/2=0.5
1/1000=0.001
1/2000000=0.0000005

Pattern: As you divide 1 by a larger number the result gets smaller. This means that the value of 1/x tends towards 0 as x approaches infinity. So we say that 1/infinity is 0.

Think of the graph of 1/x.

Hope this helps. :smile:

Thankyou!! :smile:

Reply 5

JemmaSimmons

Original post by elvss567

Thankyou!! :smile:

No problem. :smile:

Reply 6

RDKGames

Study Forum Helper

Original post by JemmaSimmons

Integrating gives -1/4*(4x+1)-1 (Agree with you.)

Therefore the value of the infinite integral is:
-1/4[1/infinity -1/1)=-1/4[0+-1]= -1/4*-1 = 1/4

Hope this helps. :smile:

Edit: I took -1/4 out as a factor. Doing it without factorising gives (-1/4*1/infinity) - (-1/4*1)=0--1/4=1/4

Original post by elvss567

how did you get zero for the infinity bit? what did you substitute in for x? is there an infinity button on the calculator or something? :smile:

Substituting infinity is a very bad practice, infinity isn't even a number you can use in these calculations!

Essentially you have

\displaystyle \int_{0}^{\infty} (4x+1)^{-2} .dx = \lim_{a \rightarrow \infty} \int_{0}^{a} (4x+1)^{-2}

\displaystyle I= \lim_{a \rightarrow \infty} -\frac{1}{4} \left[ \frac{1}{(4x+1)} \right]_0^a

Thus

Unparseable latex formula:

\displaystyle I= \lim_{a \rightarrow \infty} -\frac{1}{4} \left( \frac{1}{(4a+1)} - \frac{1}{4(0)+1}} \right)

\displaystyle \Rightarrow I=-\frac{1}{4} \left( \lim_{a \rightarrow \infty} \frac{1}{4a+1} -1 \right)

Now you need to observe what happens to that term as a goes to infinity, thus what the term's limit is as that happens, then use that value in there.

(edited 6 years ago)

Reply 7

RDKGames

Study Forum Helper

Original post by JemmaSimmons

So we say that 1/infinity is 0.

No, we say

\displaystyle \lim_{x \rightarrow \infty} \frac{1}{x} = 0

Reply 8

JemmaSimmons

Original post by RDKGames

No, we say

\displaystyle \lim_{x \rightarrow \infty} \frac{1}{x} = 0

*Except I have no idea how to type that haha, so just went with the simple, albeit not quite right, version of the explanation. I am also of the mindset that it is best not to sub in infinity as it is not actually a number.

"This means that the value of 1/x tends towards 0 as x approaches infinity. So we replace 1/infinity with 0. " <-Improved version. :smile:

(edited 6 years ago)

Reply 9

elvss567

got it! thankyou both!!