integrals...

**tennisgirl** · April 29th 2008, 04:55 PM

(1 pt) Consider the function

Evaluate the definite integral.

This is a homework problem. I got an answer of (1/2) - ln(9), but that is incorrect. I used the properties of integrals to split up the function. Can you give me a hint? Thanks!

**Mathstud28** · April 29th 2008, 05:10 PM

Originally Posted by tennisgirl

(1 pt) Consider the function

Evaluate the definite integral.

This is a homework problem. I got an answer of (1/2) - ln(9), but that is incorrect. I used the properties of integrals to split up the function. Can you give me a hint? Thanks!

If you look at it from 0<x<1...and from 1<x it is 1/x.....so the area should be
$\int_0^{1}xdx+\int_1^{9}\frac{1}{x}dx=[\frac{1}{2}-0]+[\ln(9)-\ln(1)=\frac{1}{2}+\ln(9)$

**o_O** · April 29th 2008, 05:12 PM

$\int_{0}^{9} f(x) dx = \int_{0}^{1} f(x) dx + \int_{1}^{9} f(x)dx$
$= \int_{0}^{1} x dx + \int_{1}^{9} x^{-1} dx$
$= \frac{1}{2}x^{2} \bigg|_{0}^{1} \: \: + \: \: \ln x \bigg|_{1}^{9}$

Try going through it more carefully

Ah darn, 2.5 - 2 now eh ?

**Mathstud28** · April 29th 2008, 05:14 PM

Originally Posted by Mathstud28

If you look at it from 0<x<1...and from 1<x it is 1/x.....so the area should be
$\int_0^{1}xdx+\int_1^{9}\frac{1}{x}dx=[\frac{1}{2}-0]+[\ln(9)-\ln(1)=\frac{1}{2}+\ln(9)$

Note! This applies because $\lim_{x\to{1^+}}f(x)=\lim_{x\to{1^-}}f(x)$
and now that we know that the limit exists..this is only true because

$\lim_{x\to{1}}f(x)=f(1)$

i.e. it is continous on [0,9]

**Mathstud28** · April 29th 2008, 05:22 PM

Originally Posted by o_O

$\int_{0}^{9} f(x) dx = \int_{0}^{1} f(x) dx + \int_{1}^{9} f(x)dx$
$= \int_{0}^{1} x dx + \int_{1}^{9} x^{-1} dx$
$= \frac{1}{2}x^{2} \bigg|_{0}^{1} \: \: + \: \: \ln x \bigg|_{1}^{9}$

Try going through it more carefully

Ah darn, 2.5 - 2 now eh ?

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