$\displaystyle

\int_{-2}^{-1} \frac{1}{3-4x}

$

$\displaystyle

= (\frac{-1}{4}log(3-4(-1))) - (\frac{-1}{4}log(3-4(-2)))

$

$\displaystyle

= 0.049073661

$

correct? thank you

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- Nov 10th 2008, 03:37 AMjvignaciointegration check 2
$\displaystyle

\int_{-2}^{-1} \frac{1}{3-4x}

$

$\displaystyle

= (\frac{-1}{4}log(3-4(-1))) - (\frac{-1}{4}log(3-4(-2)))

$

$\displaystyle

= 0.049073661

$

correct? thank you - Nov 10th 2008, 03:39 AMmr fantastic
- Nov 10th 2008, 04:00 AMjvignacio
- Nov 10th 2008, 04:02 AMmr fantastic
- Nov 10th 2008, 04:15 AMjvignacio
- Nov 10th 2008, 04:21 AMmr fantastic
- Nov 10th 2008, 04:28 AMjvignacio
- Nov 10th 2008, 04:50 AMqorilla
The 'log' on your calculator means logarithm to base 10.

Use the 'ln' button for e-based logarithm. - Nov 10th 2008, 06:48 AMjvignacio