Question

Calculus. Integrals

Answer 1

\[\int\limits_{0}^{1}\frac{ 29 }{ 4y-1 } dy\]

Answer 2

determine if the integral is convergent or divergent. If convergent, evaluate.

Answer 3

do you think it is a convergent integral ?

Answer 4

also, to evaluate it, I would use: \(\large\color{slate}{\displaystyle u=4y-1}\)

Answer 5

i think its convergent.... improper integral

Answer 6

i was thinking the lim would be a --> 0+

Answer 7

I got disconnected, and now back. Apologize,.

Answer 8

\(\large\color{slate}{\displaystyle\int\limits_{y{\Large=}0}^{y {\Large=}1}\frac{29}{4y-1}dy}\)
~~~~~~~~~~~~~~~~~~~~

\(\large\color{slate}{u=4y-1}\)

\(\large\color{slate}{\displaystyle \frac{1}{4}du=dx}\)

\(\large\color{slate}{y=0~~~~~\Rightarrow~~~~~u=4(0)-1\Rightarrow~~~~~u=-1}\)

\(\large\color{slate}{y=1~~~~~\Rightarrow~~~~~u=4(1)-1\Rightarrow~~~~~u=3}\)

~~~~~~~~~~~~~~~~~~~~
\(\large\color{slate}{\displaystyle\frac{29}{4}\int\limits_{u{\Large=}-1}^{u {\Large=}3}\frac{1}{u}du}\)


\(\large\color{slate}{\displaystyle\frac{29}{4}\int\limits_{u{\Large=}-1}^{u {\Large=}3}\frac{1}{u}du=\frac{29}{4}\ln|u|~~{{\huge |}_{ -1}^{3}}}\)

Answer 9

(I didn't forget to put the absolute value for the ln as well, as you can see.... )

Answer 10

So,
\(\large\color{slate}{ \ln|3|=\ln(3) }\)
\(\large\color{slate}{ \ln|-1|=\ln(1)=0 }\)

Answer 11

Again, ... sorry

Answer 12

So the answer is a real number, \(\large\color{slate}{ \displaystyle \frac{29}{4}\ln(3) }\)

Answer 13

and, yes, as you figured. It is convergent

Answer 14

@SolomonZelman a singularity exists at \(y=\dfrac{1}{4}\). We have to consider the integral in parts:
\[\large\int_0^1\frac{29}{4y-1}\,dy=\left\{\lim_{c\to1/4^-}\int_0^{c}+\lim_{c\to1/4^+}\int_c^1\right\}\frac{29}{4y-1}\,dy\]

Answer 15

srry, I think I left something out again.

Answer 16

tnx for showing.