area under the curve

• June 20th 2007, 11:18 AM
viet
area under the curve
Find the area of region under the curve $y = \frac{1}{x^2+x}$ to the right of $x = 1$
• June 20th 2007, 11:25 AM
topsquark
Quote:

Originally Posted by viet
Find the area of region under the curve $y = \frac{1}{x^2+x}$ to the right of $x = 1$

You are looking for
$\int_1^{\infty} \frac{dx}{x^2 + x}$

Hint: $x^2 + x = x^2 + x + \frac{1}{4} - \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2 - \frac{1}{4}$

Put this in the denominator and use the substitution $y = x + \frac{1}{2}$. Does this integral look more familiar?

-Dan
• June 20th 2007, 11:30 AM
curvature
Quote:

Originally Posted by topsquark
You are looking for
$\int_1^{\infty} \frac{dx}{x^2 + x}$

Hint: $x^2 + x = x^2 + x + \frac{1}{4} - \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2 - \frac{1}{4}$

Put this in the denominator and use the substitution $y = x + \frac{1}{2}$. Does this integral look more familiar?

-Dan

Notice that 1/(x^2+x)=1/x-1/(x+1)
• June 20th 2007, 11:50 AM
curvature
Quote:

Originally Posted by curvature
Notice that 1/(x^2+x)=1/x-1/(x+1)

the region looks like this
• June 20th 2007, 06:05 PM
Krizalid
Quote:
To viet: finally, you take the limit with $\infty$, then it's done.
• June 20th 2007, 06:48 PM
viet
i got the final answer to be $ln 2$. thanks y'all :)