Integration (6) Flashcards

1
Q

What to do if integration is a standard result?

A

Does it need scaling

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2
Q

What to think of when manipulating it to a standard result?

A

Think trig identities and expanding brackets

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3
Q

When can I use the reverse chain rule?

A

Is the numerator a derivative of the denominator

Is one factor related to the derivative of another

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4
Q

When can I split the numerator?

A

Is there a single term in the denominator?

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5
Q

When can I do partial fractions?

A

Does the denominator factorise?

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6
Q

When can I do algebraic division?

A

Is the fraction improper?

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7
Q

Use reverse chain rule for ∫cosxsin^2x dx

A

Consider: (sinx)^3
Differentiate: 3cosxsin^2x
Scaling: We want 1/3

(1/3)sin^3x + C

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8
Q

Use reverse chain rule for ∫x(x^2+5x)^3

A

Consider: (x^2+5x)^4
Differentiate: 8x(x^2+5x)^3
Scaling we want: (1/8)

(1/8)(x^2+5x)^4 + C

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9
Q

Use reverse chain rule for ∫(2x)/(x^2+1)

A

Consider: ln|x^2 + 1|
Differentiate: (2x)/(x^2+1)
Scaling we want: None required

ln|x^2 + 1| + C

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10
Q

Use u=2x+5 to find ∫x¬(2x+5) dx

A

Find x:
u = 2x+5
x = (u-5)/2

Find dx:
u = 2x+5
du/dx = 2
dx = (1/2)du

Substitute these into expression:
∫((u-5)/2) * u^(1/2) * (1/2)du
∫((u-5) * u^(1/2))/4 du
(1/4) ∫(u-5) * u^(1/2) du
(1/4) ∫u^(3/2) - 5u^(1/2)

Integrate simplified expression:
(1/4) (2/5)u^(5/2) - (10/3)u^(3/2) + C
(1/10)u^(5/2) - (5/6)u^(3/2) + C

Write answer in terns of x:
Replace u with (2x+5)

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11
Q

Calculate (pi/2)∫0 (cosx)¬(1+sinx) dx

A
u = 1 + sinx
du/dx = cosx
dx = (1/cosx) du 

Sub these values into: u = 1 + sinx
x = pi/2, u = 2
x = 0, u = 1

2∫1 cosx * u^(1/2) * (1/cosx) du
2∫1 u^(1/2) du

= 2[(2/3)u^(3/2)]1
= (2/3)2^(3/2) - (2/3)1^(3/2)

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12
Q

Find the general solution to (1+x^2)(dy/dx) = xtany

A
(1+x^2)(dy/dx) = xtany
(1/tany)(dy/dx) = x/(1+x^2)

∫ (1/tany) dy = ∫ x/(1+x^2) dx
∫ coty dy = ∫ x/(1+x^2) dx

ln|siny| = (1/2)ln|1+x^2| + lnk
ln|siny| = lnk¬(1+x^2)
siny = k¬(1+x^2)
y = arcsin(k¬(1+x^2))
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13
Q

The rate of increase of a rabbit population (with population P, where time is t) is proportional to the current population. Form a differential equation and find its general solution

A

dp/dt = kP
(1/P) dp/dt = k

∫(1/P) dP = ∫k dt
ln|P| = kt + C
P = e^(kt + c)
P = Ae^(kt)

t=0, p=300
300 = Ae^0
A = 300

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14
Q

Determine the area bound between the curve with parametric equations x = t^2 and y = t + 1, the x-axis, and the lines x = 0 and x = 3

A

∫y (dx/dt) dt

x = t^2
dx/dt = 2t
y = t + 1
x = 3, t = ¬3
x = 0, t = 0
¬3∫0 (t + 1)2t dt
¬3∫0 (2t^2 + 2t) dt
= ¬3[(2/3)t^3 + t^2]0
= (2/3)*(¬3)^3 + (¬3)^2
= 2¬3 + 3
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