study sheet -- yeah -- >wink wink<
still, what the hell.
for the first one, these are pretty standard integral forms
β«[2,5] x^2 dx = 1/3 x^3 [2,5] = 1/3 (5^3-2^3)
if u = 2x, then du = 2 dx, so
β«[0,Ο/2] 2sin(2x) dx = β«[0,Ο] sin(u) du = -cos(u) [0,Ο] = -(-1 - 1) = 2
if u = x^3-1, then du = 3x^2 dx
β«[2,4] 3x^2 β(x^3-1) dx = β«[7,65] βu du = 2/3 u^(3/2) = 2/3 (65β65 - 7β7)
recall that if F(t) = β«[u,v] f(x) dx, then d/dt F(t) = f(v) dv/dt - f(u) du/dt
This is just the chain rule in reverse.
β«[t^2,t^3] x^2 dx = (t^3)^2 (3t^2) - (t^2)^2 (2t) = 3t^8 - 2t^5
If u = sin(2x) then du = 2cos(2x) dx
so, β«[Ο/4,Ο/2] 2sin(2x) cos(2x) dx = β«[1,0] u du = 1/2 u^2 [1,0] = -1/2
or,
β«[Ο/4,Ο/2] 2sin(2x) cos(2x) dx = β«[Ο/4,Ο/2] sin(4x) dx
= -1/4 cos(4x) [Ο/4,Ο/2] = -1/4 (1 - -1) = -1/2
for the 2nd one,
If F(x) = β« f(x) dx
β«[a,b] f(x) dx = F(b) - F(a)
β«[a,b] f(x+a) dx = F(a+b)-F(2a)
β«[a-c,b-c] f(x+c) dx = F(b-c+c)-F(a-c+c) = F(b)-F(a)
h'(x) = tan^-1(x)
β«[1,5] f'(x) dx = f(5)-f(1) = 9
f(5)-11 = 9
...
F(x) = β«[cosx,sinx] βt dt
F'(x) = βsinx (cosx) - βcosx (-sinx)
v(h) = β«[1,h] Ο/y^2 dy
dv/dt = dv/dh * dh/dt = Ο/h^2 dh/dt
3/2 = Ο/4 dh/dt
...