Integration (4) Flashcards
∫25 + 20x^(1/2) + 4x
25x + (40/3)x^(3/2) + 2x^2 + C
25x + (40/3)x^(3/2) + 2x^2 + C
∫25 + 20x^(1/2) + 4x
Curve C has the derivative f’(x) = x^3 - 2x^-2 . The point P (1,2) lies on C.
Find f(x)
Integrate f’(x)
Substitute x and y into the integral
Find C
State f(x)
The diagram on the left shows the curve y = (x+1)(x-5) at points J (-1,0) and K (4, -5) lie on the curve.
Find the equation of the straight line joining J and K in the form y = mx+c
m = (0–5)/(-1-4) = -1
y - (-5) = -1(x - 4) => y + 5 = -x + 4
y = -x - 1
The diagram on the left shows the curve y = (x+1)(x-5) at points J (-1,0) and K (4, -5) lie on the curve.
Calculate 4∫-1 (x+1)(x-5)dx
x^2 - 4x - 5
(1/3)x^3 - 2x^2 - 5x + C
[(1/3)(4)^3 - 2(4)^2 - 5(4)] - [(1/3)(-1)^3 - 2(-1)^2 - 5(-1)] =
-30(2/3) - 2(2/3) = -33(1/3)
Integrating a parametric
∫ y dx/dt dt
∫ y dx/dt dt
Integrating a parametric
The curve C on the right has parametric equations x = t^2 + 3t and y = t^2 + (1/t^3), t>0. The shaded region marked R is enclosed by C, the x-axis and the lines x=4 and x=18
The area R = 18∫4 y dx
dx/dt = 2t + 3
Change the limits of the integral
x = 18 => t^2 + 3t - 18 = 0 => (t-3)(t+6)=0 => t= 3 or -6
x = 4 => t^2 + 3t - 4 = 0 => (t-1)(t+4)=0 => t= 1 or -4
The limits are 3 and 1 as t>0
R = 3∫1 y (dx/dt) dt = 3∫1 (t^2 + (1/t^3)) * (2t + 3) dt
= 3∫1 ((t^5 + 1)/t^3)) * (2t + 3)
= 3∫1 ((t^5 + 1)(2t+3))/(t^3)
Integrate 5/x
∫ 5/x dx = 5 ∫ 1/x dx = 5 ln|x| + C
Integrate 1/(4x+5)
(1/4) ln|4x+5| + C
Find ∫ (3 sin 3x)/(cos 3x + 2)
d/dx (cos 3x + 2) = -3 sin 3x
As the numerator is -1 * the derivative of the denominator. So
∫(3 sin 3x)/(cos 3x + 2) = - ∫-(3 sin 3x)/(cos 3x + 2)
= -ln|cos 3x + 2| + C
Find ∫ 2e^(4-3x) dx
-(2/3)e^(4-3x) + C
-(2/3)e^(4-3x) + C
Find ∫ 2e^(4-3x) dx
Integrate -2 sin 2x dx
-2 (-(1/2) cos 2x) = cos 2x
-2 (-(1/2) cos 2x) = cos 2x
Integrate -2 sin 2x dx
