Hello!! I'm trying to create a study sheet to help myself! If someone could give and explain the correct answers, It would be very helpful for me and appreciated!
gyazo.com/0dccabaac7b9ed649b06d394ce306ff5
gyazo.com/471e9983d44105f9673e437c6ab56609
(sorry for the links, I thought it would be easier to see them as images, because I'm not sure how to type them out on here!)
Again, thank you so much for taking time to view my question and possibly help me!!
study sheet -- yeah -- >wink wink<
still, what the hell.
for the first one, these are pretty standard integral forms
∫[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
∫[0,π/2] 2sin(2x) dx = ∫[0,π] sin(u) du = -cos(u) [0,π] = -(-1 - 1) = 2
if u = x^3-1, then du = 3x^2 dx
∫[2,4] 3x^2 √(x^3-1) dx = ∫[7,65] √u du = 2/3 u^(3/2) = 2/3 (65√65 - 7√7)
recall that if F(t) = ∫[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.
∫[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, ∫[π/4,π/2] 2sin(2x) cos(2x) dx = ∫[1,0] u du = 1/2 u^2 [1,0] = -1/2
or,
∫[π/4,π/2] 2sin(2x) cos(2x) dx = ∫[π/4,π/2] sin(4x) dx
= -1/4 cos(4x) [π/4,π/2] = -1/4 (1 - -1) = -1/2
for the 2nd one,
If F(x) = ∫ f(x) dx
∫[a,b] f(x) dx = F(b) - F(a)
∫[a,b] f(x+a) dx = F(a+b)-F(2a)
∫[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)
∫[1,5] f'(x) dx = f(5)-f(1) = 9
f(5)-11 = 9
...
F(x) = ∫[cosx,sinx] √t dt
F'(x) = √sinx (cosx) - √cosx (-sinx)
v(h) = ∫[1,h] π/y^2 dy
dv/dt = dv/dh * dh/dt = π/h^2 dh/dt
3/2 = π/4 dh/dt
...