Difference between revisions of "Integration By Parts"

Latest revision as of 06:16, 17 September 2021

Performing Integration By Parts $\int f(x)g'(x)\,dx=f(x)g(x)-\int f'(x)g(x)\,dx$

When you are looking at this equation

$\int u\,dv=uv-\int v\,du$

you may have been confused by u,du,v,dv

you can just under stand it as

so by looking at u and v as a function we you say du or dv is derivative of that function.

for example if we say u is f(x) than du is derivative of f(x) so it will be f'(x).

$\int u\,dv=uv-\int v\,du$

can be think as

$\int f(x)\,g'(x)=(f(x)*g(x))-\int g(x)\,f'(x)$

@@ Line 1: / Line 1: @@
+<noinclude>
 ====[[Integration By Parts]]====
-#Performing Integration By Parts #<math>\int f(x) g'(x) \,dx = f(x)g(x) - \int f'(x) g(x) \,dx</math>
+</noinclude>
+#Performing Integration By Parts <math>\int f(x) g'(x) \,dx = f(x)g(x) - \int f'(x) g(x) \,dx </math>
+#Performing Integration By Parts <math>\int u \,dv = uv - \int v \,du</math>
+<noinclude>
+====Explaining====
+When you are looking at this equation
+<math>\int u \,dv = uv - \int v \,du</math>
+you may have been confused by
+u,du,v,dv
+you can just under stand it as
+#u = f(x)
+#du = f'(x)
+#v = g(x)
+#dv = g'(x)
+so by looking at u and v as a function we you say du or dv is derivative of that function.
+for example if we say u is f(x) than du is derivative of f(x) so it will be f'(x).
+<math>\int u \,dv = uv - \int v \,du</math>
+can be think as
+<math>\int f(x) \,g'(x) = (f(x) * g(x)) - \int g(x) \, f'(x)</math>
+</noinclude>