From f0a0437612477cb3eb353f3238c869abb9e6e04a Mon Sep 17 00:00:00 2001 From: arc Date: Wed, 3 Sep 2025 12:30:04 -0600 Subject: [PATCH] vault backup: 2025-09-03 12:30:04 --- .../math/MATH1220 (calc II)/Integration by Parts.md | 11 ++++++++++- 1 file changed, 10 insertions(+), 1 deletion(-) diff --git a/education/math/MATH1220 (calc II)/Integration by Parts.md b/education/math/MATH1220 (calc II)/Integration by Parts.md index 9bcfe2b..477fe87 100644 --- a/education/math/MATH1220 (calc II)/Integration by Parts.md +++ b/education/math/MATH1220 (calc II)/Integration by Parts.md @@ -23,4 +23,13 @@ $$\int xe^{2x}dx$$ - $dv = e^{2x}dx$ - $v = \frac{1}{2}e^{2x}$ - The antiderivative of $dv$. 2. Looking back at the integration by parts formula, we know that: - $$ \int udv = uv - \int v du $$ \ No newline at end of file + $$ \int udv = uv - \int v du $$ + $$ \int xe^{2x}dx = (\frac{1}{2}e^{2x})(x)-\int (\frac{1}{2}e^{2x}) (1dx) $$ +3. The remaining integral can be solved with $u$ substitution, but we've already defined $u$, so we use $w$ as a replacement. + - $w = 2x$ + - $dw = 2dx$ + - $\frac{1}{2}dw=dx$ + 1. Substituting $w$ and $dw$ into the integral: + $$ \int \frac{1}{2}e^w \frac{1}{2}dw $$ + 2. This gives an integral that can be computed naively + $$ \int\frac{1}{2}e^{w} $$ \ No newline at end of file