Question

Find the area of the region bounded by
y = 9 / (25 − x^2)
and y = 1.

Answer 1

help, please?

Answer 2

I have gotten the same answer as wolfram alpha, and I don't know what is wrong

Answer 3

\[x + 9/10 \ln(5-x) - 9/10\ln(x+5) \]

Answer 4

our bounds are from -4 to 4 right?

Answer 5

i don't have bounds for this question.

Answer 6

you do have bounds; you just got to determine their value; they are where the lines cross

Answer 7

wait, what do you mean?

Answer 8

oh... I need to put the bounds into the equation and solve for an number as the answer

Answer 9

we can do this simpler by taking the bounds from 0 to 4 and doubling it since they are equal on both sides of the y axis

Answer 10

the area of 4 wide and 1 tall rectangle formed by y=1 and x=4 is an area of 4 :) now subtract away that area under the 9/(25-x^2)

Answer 11

4-9/10(ln(9))

Answer 12

if the integration is good; then yes; now double it to get both areas on the left and right of the yaxis

Answer 13

multiply by 2....4 - 9/5(ln(9))? but is says it is wrong...

Answer 14

\[\int\frac{1}{a^2-x^2}dx\implies\ \frac{1}{2a}\ ln\left|\frac{a+x}{a-x}\right|\]

Answer 15

a=5; x=x in this case; then bring back the 9 right?

Answer 16

yep

Answer 17

let me right it again in the system....this is right

Answer 18

\[\frac{9}{10}ln\left|\frac{5-4}{5+4}\right|-\frac{9}{10}ln\left|\frac{5}{5}\right|\]

Answer 19

\[2\ [\ 4-\frac{9}{10}\left(ln(1/9)\right)]\] is what I get if I did it right :)

Answer 20

how did you get 9 on the bottom?

Answer 21

a = 5; and x=4 it the bounds... just plugged it in

Answer 22

and I got it upside down too lol

Answer 23

oh...fraction of ln

Answer 24

wait.....ln|9| - ln|1| = 1/9?

Answer 25

\[8 - \frac{9\ ln(9)}{5}\]

Answer 26

i had my 5+x and my 5-x transposed

Answer 27

oh...kay....now it is right. What I didn't do is multiply 4 *2, I just divided by 2

Answer 28

yeah you did, but you got it right. thanks for the help!

Answer 29

youre welcome :)