∫2xcos(x2)dx
sin(x2)+C
the volume when y=x2, from 0 to 1, rotated about the x-axis.
π/5
When should you use shells instead of washers?
when slicing parallel to axis of rotation or when solving for the other variable is messy.(draw picture)
Hooke's Law (spring) F=
F=kx
Derivative of lnx
∫exdx
1/x
ex+c
∫1/(xlnx)dx
ln∣lnx∣+C
the volume when y=x2, from 0 to 1, rotated about the y-axis
π/2
Rotate y=x2, 0→1, about y-axis (shells)
π/2
Spring stretched 2m with force 10N → find k
5
Derivative of e3x
∫1/xdx
3e 3x
ln∣x∣+C
∫xe(x^2)dx
1/2e^x^2+C
Region between y=x and y=x2, rotated about x-axis
2π /15
shell setup
V=2π∫(radius)(height)dx
Work to stretch a spring (k=5) from 1m to 3m
20
(f−1)′(a)
∫e2xdx
1/(f′(f−1(a)))
1/2e2x+C
∫x/(sqrt(1+x2))dx
sqrt(1+x^2)+C
the washer setup
& the washer setup key
V=π∫(R2−r2)dx
outer radius − inner radius
slices impact
A rope of length 10 m hangs over the side of a building. The rope has a mass density of 2 kg/m. How much work is required to pull the entire rope up to the top?
980 J
Differentiate: y=ln(x2+1)
2x/(x2+1)
∫sin3xcosxdx
(sin4x)/4 +C
Switch to x=sqrt(y)
V=π∫(R2−r2)dy
Compare slicing vs shells; which is easier and why?
A bucket containing 20 kg of water is lifted 10 m to the top of a building. The bucket itself weighs 5 kg. As it is lifted, water leaks out at a constant rate, so that by the time it reaches the top, the bucket is empty. How much work is done?
1470 J
Differentiate: y=xx
xx(lnx+1)