Thread: integral problem

This problem has me stumped:

Find the exact area of the region:

Bounded by y=(1)/(sqrt(x^2+9)), y=0, x=0, x=3

I'm a little rusty on trigonometric substitutions

Originally Posted by Nitz456

This problem has me stumped:

Find the exact area of the region:

Bounded by y=(1)/(sqrt(x^2+9)), y=0, x=0, x=3

I'm a little rusty on trigonometric substitutions

Are you wanting:

0 = int(1/(sqrt(x^2 + 9))dx), from x = 0...3, in which case this is not possible.

The integral is ln(sqrt(2) + 1), and this will never equal 0.

Originally Posted by AfterShock

Are you wanting:

0 = int(1/(sqrt(x^2 + 9))dx), from x = 0...3, in which case this is not possible.

The integral is ln(sqrt(2) + 1), and this will never equal 0.

What I'm looking for, I'm quite sure, is the area of the integral of the equation i was given, for 0..3.

i think the y=0 is just meaning that it doesnt want the integral below the x-axis.