How do I find the exact area of the region bounded by y=1/sqrt(x^2 + 9), y=0, x=0, x=3?

Answer 1

The region is given by the following integral:
3
∫ 1/√(x² + 9) dx
0

You can solve this integral by substitution
set x = 3·sinh(z) ←→ z = arsinh(x/3)
→ dx = 3·cosh(z)
∫ 1/√(x² + 9) dx
= ∫ 3·cosh(z) / √(9·sinh²(z) + 9) dz
= ∫ cosh(z) / √(sinh²(z) + 1) dz
because cosh²(z) - sinh²(z) = 1
= ∫ √(sinh²(z) + 1) / √(sinh²(z) + 1) dz
= ∫ 1 dz
= z + c
= arsinh(x/3) + c

Therefore:
3
∫ 1/√(x² + 9) dx
0
= arsinh(1) - arsinh(0)
= arsinh(1)
because arsinh(z) = ln(z + √(z² + 1))
= ln(1 + √2)
= .88137...

Answer 2

Okay, here it is given that the boundary is from 0 to 3.
Integrate the equation within the boundary:
Integrate from 0 (lower case)to 3(upper case) 1/sqrt(x^2 + 9) =
(x^2 + 9)^(1/2)/ (1/2) from 0 to 3 = 2(x^2+9)^(1/2) from 0 to 3 = 2(sqrt 18 - sqrt 9)= 2sqrt18 - 6

note: 1/ sqrt (x^2+9) = (x^2+9)^(-1/2)

Answer 3

Integrate the function with limits of 3 and 0.
The function integrates as natural log (x^2+x+9).
At 3, the value is ln 21.
At 0. the value is ln 9.
So the value is ln (21/9) or ln (7/3).

Answer 4

why u r troubling with quantity ? basically remedy it via integration initially locate factors on graph as follows :on line y=secx while x=0 ,y=a million,while y=0, x=pi/2 subsequently factors is (0,a million) and (pi/2,0) while x=pi/3, then y =2 subsequently factor is (pi/3,2) now remedy it graphically its effortless if any problem u can use actual graph paper, it wil be much less stressful

Answer 5

good luck...