The Antiderivative of 3e^-x: Could someone please explain what it is and why? Thanks.
L Lime New member Joined Sep 8, 2006 Messages 49 Nov 12, 2006 #1 The Antiderivative of 3e^-x: Could someone please explain what it is and why? Thanks.
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Nov 12, 2006 #2 \(\displaystyle \L \int 3e^{-x} dx\) \(\displaystyle \L -3 \int e^{-x} (-dx)\) let u = -x, du = -dx ... substitute \(\displaystyle \L -3 \int e^u du = -3e^u + C\) back substitute ... \(\displaystyle \L -3e^{-x} + C\)
\(\displaystyle \L \int 3e^{-x} dx\) \(\displaystyle \L -3 \int e^{-x} (-dx)\) let u = -x, du = -dx ... substitute \(\displaystyle \L -3 \int e^u du = -3e^u + C\) back substitute ... \(\displaystyle \L -3e^{-x} + C\)
L Lime New member Joined Sep 8, 2006 Messages 49 Nov 12, 2006 #3 On the 2nd step I noticed you multiplied the 3e^-x by -1 and the dx by -1. Why?
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Nov 13, 2006 #4 I was setting up the integrand for the substitution u = -x and du = -dx.
