# area under the curve

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

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

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

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

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

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

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

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

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

-Dan

Notice that 1/(x^2+x)=1/x-1/(x+1)
• Jun 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
• Jun 20th 2007, 06:05 PM
Krizalid
Quote:
To viet: finally, you take the limit with $\displaystyle \infty$, then it's done.
• Jun 20th 2007, 06:48 PM
viet
i got the final answer to be $\displaystyle ln 2$. thanks y'all :)