# Thread: Integration - Find the area

1. ## Integration - Find the area

1. Find the area of the finite region bounded by the curve with equation y = coth x, the x-axis and the lines x =2 and x = 4.

2. The finite region R is bounded by the curves with equations y = cosh x, y = sinh x. the y-axis and the line x=1. Find the area of R.

2. Originally Posted by geton
1. Find the area of the finite region bounded by the curve with equation y = coth x, the x-axis and the lines x =2 and x = 4.

2. The finite region R is bounded by the curves with equations y = cosh x, y = sinh x. the y-axis and the line x=1. Find the area of R.
Where are you stuck? Setting up the integrals? Solving the integrals?

3. Did you draw the graphs to see what is going on?

Originally Posted by geton
1. Find the area of the finite region bounded by the curve with equation y = coth x, the x-axis and the lines x =2 and x = 4.
$\displaystyle \int_2^4 \coth x~dx$

2. The finite region R is bounded by the curves with equations y = cosh x, y = sinh x. the y-axis and the line x=1. Find the area of R.
$\displaystyle \int_0^1 (\cosh x - \sinh x)~dx$

4. Originally Posted by mr fantastic
Where are you stuck? Setting up the integrals? Solving the integrals?

For 1,

Use the identity coth x = cosh x/sinh x = (sinh x cosh x)/(sinh^2 x)

But sinh^2 x ≡ cosh^2 x – 1
So we’ve coth x = (sinh x cosh x)/ (cosh^2 x - 1) = ½ sinh 2x /( cosh^2 x - 1)

If I write cosh x = t, then sinh dx/dt = 1

From this, I’ve great problem.

5. Originally Posted by geton
For 1,

Use the identity coth x = cosh x/sinh x ...
from here, why not just make a substitution u = sinh x? why are you killing yourself?

6. Originally Posted by Jhevon
Did you draw the graphs to see what is going on?
I didn't draw the graph & it is one of my great problem.

$\displaystyle \int_0^1 (\cosh x - \sinh x)~dx$
Thank you for this.