Multi
Iterated Integrals
- Work from the inside out
- Partial Derivative but backwards
- Desmos Got that
Surface Integrals
Switching order of integration
Integration by parts
∫f(x)∗g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx
- Find an f(x) where deriving simplifies it
- Find a g’(x) where anti deriving doesn’t make it more complicated
Simple Example
∫0πxcos(x)dx
- x becomes simpler when derived, so is f(x)
f(x)=x
f′(x)=1
- cos(x) anti derivative doesn’t get more complicated, so it’s g(x)
g′(x)=sin(x)
g(x)=sin(x)
- And so we get things which we can substitute into the formula
x∗sin(x)−∫1sin(x)dx
x∗sin(x)+∫−sin(x)dx
x∗sin(x)+cos(x)dx
x∗sin(x)+cos(x)∣0π
More Examples
∫1ex2−2ln(x)
f(x)=−2ln(x)
f′(x)=x−2
g′(x)=x21
g(x)=−x1
−2ln(x)∗−x1−∫x−2∗−x1dx
−2ln(x)∗−x1+x2
−2ln(x)∗−x1+x2∣1e
∫7xsin(x)
f(x)=7x
f′(x)=7
g′(x)=sin(x)
g(x)=−cos(x)
7x∗−cos(x)−∫7∗−cos(x)
7x∗−cos(x)+∫7∗cos(x)
−7xcos(x)+7sin(x)
∫14t2ln(5t)dt
f(t)=ln(5t)
f′(t)=1/t
g′(t)=t2
g(x)=3t3
ln(5t)3t3−∫t13t3dx
ln(5t)3t3−∫3t2dx
ln(5t)3t3−9t3
∫12r3ln(r)
f(x)=ln(r)
f′(x)=1/x
g′(x)=r−3
g(x)=−2r−2
ln(r)−2r−2−∫x1−2r−2
ln(r)−2r−2−∫−2xr−2
ln(r)−2r−2−4r21
Maybe wrong skull
ln(x)x11
f(x)=ln(x)
f′(x)=1/x
g′(x)=x11/2
g(x)=x9/2/(11/2)
?????