Voortgezette integraalrekening - Hoofdstuk K - Wiskunde B - Getal & Ruimte - VWO5 en VWO6

Hoofdstuk K – Voortgezette integraalrekening
K.1
De substitutiemethode f(x) = sin(x) f’(x) = cos(x)
= omgekeerde kettingregel f(x) = cos(x) f’(x) = –sin(x)
Voorbeeld 1: f(x) = tan(x) f’(x) = of f’(x) = 1 + tan²(x)
𝑐𝑜𝑠²𝑥
f(x) = tan(5x) = = sin(5x)
F(x) = –1/5 ln|cos(5x)| + C f(x) = sin(x) F(x) = –cos(x) + C
f(x) = cos(x) F(x) = sin(x) + C
F’(x) = –1/5 –sin(5x) 5
f(x) = tan(x) F(x) = –ln|cos(x)| + C
F’(x) = sin(5x) = tan(5x) f(x) = sin(ax+b) F(x) = – cos(ax+b) + C
𝑎

Dus F(x) = –1/5 ln|cos(5x)| + C f(x) = cos(ax+b) F(x) = sin(ax+b) + C
𝑎
f(x) = tan(ax+b) F(x) = – ln|cos(ax+b)| + C
𝑎
Voorbeeld 2:
f(x) = =

F(x) = +C
F(x) = +C
F’(x) = = =
Dus F(x) = +C

Voorbeeld 3: cos2(x) + sin2(x) = 1
f(x) = 4 cos3(x) = 4 cos(x) cos2(x) = 4 cos(x) (1 – sin2(x))
f(x) = 4 cos(x) – 4 cos(x) sin2(x) sin(2A) = 2 sin(A) ∙ cos(A)
F(x) = 4 sin(x) – sin3(x) 4 cos(2A) = cos²(A) – sin²(A)
F’(x) = 4 cos(x) – sin2(x) cos(x) 4 cos(2A) = 2 cos²(A) – 1
Dus F(x) = 4 sin(x) – sin3(x) + C cos(2A) = 1 – 2 sin²(A)




𝑎
f(x) = a f’(x) = 0 f(x) = axn F(x) = ∙ xn + 1 + C
𝑛+
f(x) = axn f’(x) = naxn–1 𝑔𝑥
f(x) = gx F(x) = +C
f(x) = c ∙ g(x) f’(x) = c ∙ g’(x) l 𝑔

f(x) = g(x) + h(x) f(x) = g’(x) + h’(x) f(x) = ex F(x) = ex + C
p(x) = f(x) ∙ g(x) p’(x) = f’(x) ∙ g(x) + f(x) ∙ g’(x) f(x) = F(x) = ln|x| + C
𝑥
𝑡 𝑥 𝑛 𝑥 𝑡 ′ 𝑥 − 𝑛′ 𝑥 𝑡 𝑥
q(x) = q’(x) = f(x) = ln(x) F(x) = x ln(x) – x + C
𝑛 𝑥 𝑛 𝑥
𝑔
f(x) = gx f’(x) = gx ∙ ln(g) f(x) = ⬚log 𝑥 F(x) = ∙ (x ln(x) – x) + C
l 𝑔

f(x) = ln(x) f’(x) = 𝑥
𝑛
f(x) = ln(xn) f’(x) = 𝑥
𝑔
f(x) = ⬚log 𝑥 f’(x) = 𝑥 l 𝑔