Question

Check .... starting on definite integrals

Answer 1

I keep getting disconnected

Answer 2

\[\large\color{midnightblue}{ \int\limits_{1}^{5} x^5+2x-1 }\] I find each term separately, \[\large\color{midnightblue}{ \frac{x^5}{5}+x^2-x+C }\]

Now from 5 to 1.

\[\large\color{midnightblue}{ \frac{(5)^5}{5}+(5)^2-(5)+C ~~~⇒~~~645+C }\]\[\large\color{midnightblue}{ \frac{(1)^5}{5}+(1)^2-(1)+C ~~~⇒~~~ \frac{1}{5}+C }\]

ANSWER: \(\LARGE\color{black}{ 644\frac{1}{5} }\)

Answer 3

I mean 644 4/5

Answer 4

One more. Please message if I am wrong.

Answer 5

I forgot dx for the last one

Answer 6

Don't reply please, not yet

Answer 7

i think the first step\[\int\limits_{}^{}x^5=\frac{ x^6 }{ 6 }+c\]

Answer 8

yes

Answer 9

My bad, I made such a horrible errror

Answer 10

yeah silly mistake :)

Answer 11

625 × 5 = 3125

3125 +20 - 5

3140

and subtracting the sum, (3140 +C)- (1/5+C) = 3139 4/5

fixed.

Answer 12

I think I am read to move to the next one.

Answer 13

But the process is right ~

Answer 14

And you only need the +C on the indefinite..

Answer 15

yeah

Answer 16

Wait, for the first problem, it doesn't really matter. I read though that Cs DO cancel.

Answer 17

The Cs cancel when subtracting the sums, or not ?

Answer 18

yeah C will be cancelled out for definite integrals but for indefinite C will remain if there are'nt any boundary conditions

Answer 19

thanks for the clarification. Moving on....

Answer 20

\[\large\color{midnightblue}{ \int\limits_{0}^{2} ~~\frac{x^2+1}{x^2-1} ~dx } \]\[\large\color{midnightblue}{ \int\limits_{ }^{ } ~~\frac{x^2+1}{x^2-1} ~dx } \] then, I am doing long division,
\[\large\color{midnightblue}{ \int\limits_{ }^{ } ~~(-\frac{1}{x+1} +\frac{1}{x-1} +1)~dx } \]
u=x+1 du=dx
s=x-1 ds=dx

\[\large\color{midnightblue}{ -\log(u) + \int\limits_{ }^{ } \frac{1}{x-1} ~dx +x } \]

\[\large\color{midnightblue}{ -\log(u) + \log(s) +x } \]

\[\large\color{midnightblue}{ -\log(x+1) + \log(x-1) +x+C } \]

Answer 21

THen I am plugging in and subtracting the sums

log(x-1) - log(x+1) + x + C
log[ (x-1)/(x+1) ] + x + C
log[ (x-1)/(x+1) ] + x log10 + C
log[ (x-1)/(x+1) ] + log10^x + C
log (x-1)10^x /(x+1) + C

hold on is this necessary ? or should I just plug it in right away ?

Answer 22

log(x-1) - log(x+1) + x + C
log(2-1) - log(2+1) + 2 + C
log(1) - log(3) + 2 + C
2 - log(3) + C
2log(10) - log(3) + C
log(10^2) - log(3) + C
log(100/3) + C


and at x=-1 I get undefined values, log(-1) -:(

Answer 23

I mean at x=0 *

Answer 24

yeah u can simplify it a little then put the limits
\[I=[\log \frac{ x-1 }{ x+1 }+x+c]_{x=0 ~~\to~~x=2}\]

Answer 25

but what when I get an undefined values at x=0. Because I do.
What would I put as an answer ?

Answer 26

wait i'm getting 2-log(3)

Answer 27

yes, I got the same, but I did...
2-log(3)
2log(10)-log(3)
log(10²)-log(3)
log(100)-log(3)
log(100/3)

Answer 28

this is for x=2, but at x=0 I a getting undefined value. i.e. log(-1)

Answer 29

then we have to conclude that the integral does not converge

Answer 30

Okay, thank you :)

Answer 31

or u can take
log(-1)=log(1)+i*pi
using this--\[\log(-a)=\log(a)+i \pi\]

Answer 32

wait, when I plug 0 for x, I get
log(x-1) - log(x+1) + x + C
log(-1) - log(1) + x + C Using log(a)-log(b)=log(a ÷ b)
log(-1) + x + C

but I would just conclude it doesn't exist, because it is either negative log or an imaginary (?),

Answer 33

I didn't know about \(\normalsize\color{black}{ \log(-a)=\log(a)+iπ}\)

Answer 34

ty!

Answer 35

yeah this property is true for only a>0