$\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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Originally Posted by jvignacio $\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 No. The final numerical answer is wrong.
Originally Posted by mr fantastic No. The final numerical answer is wrong. the decimal places? is the substitution correct? where i enter -1 and -2 in for the x's?
Originally Posted by jvignacio the decimal places? is the substitution correct? where i enter -1 and -2 in for the x's? Originally Posted by jvignacio $\displaystyle \int_{-2}^{-1} \frac{1}{3-4x} $ $\displaystyle = (\frac{-1}{4}log(3-4(-1))) - (\frac{-1}{4}log(3-4(-2))) $ Mr F says: Correct. = 0.049073661 Mr F says: Wrong. correct? thank you ..
Originally Posted by mr fantastic .. wierd. my calculator is giving me that answer.
Originally Posted by jvignacio wierd. my calculator is giving me that answer. Then you're not entering the expression into your calculator correctly. Like the old saying says: Garbage in, garbage out.
Originally Posted by mr fantastic Then you're not entering the expression into your calculator correctly. Like the old saying says: Garbage in, garbage out. hrmmm could be. ive tryed a few things and i keep getting the same answer.. ill keep trying
The 'log' on your calculator means logarithm to base 10. Use the 'ln' button for e-based logarithm.
Originally Posted by qorilla The 'log' on your calculator means logarithm to base 10. Use the 'ln' button for e-based logarithm. much appreciated! thank you
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