Integrals Flashcards
Indefinite integral (∫____ dx)
Find the antiderivative of the ‘integrand’ contained
Integrand
The argument that is to be turned into the integral
Indefinite integrals to of a power
∫x^p dx=[x^(p+1)]/[p+1]+C
Quantity x raised to one an additional power over the entire exponent
Plus a constant
Integrals for functions multiplied by a constant
∫ c*f(x) dx=c(∫f(x) dx)
Same as that constant multiplied by the integral of that function
Integral of sums
∫ f(x)+g(x) dx=(∫f(x)dx)+(∫g(x)dx)
The integral is the sum of the integrals of each term
Integral of cos(ax)
∫cos(ax)dx= [sin(ax)]/a+C
Switched to sin and divided by ‘a’
Plus a constant
Integral of sin(ax)
∫cos(ax)dx= -[cos(ax)]/a+C
Switched to cos and divided by ‘-a’
Plus a constant
Integral of sec^2(ax)
∫sec^2(ax)dx= [tan(ax)]/a+C
Becomes tan(ax), and divided by a Plus the constant
Integral of csc^2(ax)
∫csc^2(ax)dx= -[cot(ax)]/a+C
Becomes cot(ax), and divided by '-a' Plus the constant
Integral e^(ax)
∫e^(ax)dx= [e^(ax)]/a+C
Same thing just divided by ‘a’
Plus a constant
Integral 1/x
∫(1/x)dx=ln|x|+C
Natural log (x) Plus a constant
Family of functions
Same integral but the constant varies
Go up and down on the y-axis
Integrals of acceleration
∫a(t)=v(t) Velocity
∫∫a(t)=s(t) Position
Integal of velocity
∫v(t)=s(t)
Position
Regular partition
Integration method whereby the domain is divided into equally spaced rectangles, the midpoint of each reaches up to the output on the curve
Reimann sum
Sum of the areas of all regular partition rectangles
Σ(f(x)Δx)=(f(x1)Δx)+…+…
On the interval [a,b]
Left Reimann-Sum
Sum rectangles from the upper-left corner
Underestimates on positive intervals
Overestimates on negative intervals
n-measurements start at f(0) and increase by units of Δx
Σ(f(n)*Δx)
Right Reimann Sum
Sum rectangles from the upper-right corner
Overestimates on positive intervals
Underestimates on negative intervals
n-measurements start at f(Δx) and continue to increase by units of Δx
Σ(f(n)*Δx)
Midpoint Reimann Sum
Sum rectangles from the middle of the interval
Neither overestimates or underestimates on either interval
n-measurements start at f(Δx.5) and continue to increase by units of Δx
Σ(f(n)Δx)
Definite integral
When n-measurements start at f(Δx.5) and continue to increase by units of Δx
Lim as Δx→0 Σ(f(n)Δx)=∫f(x)
Or
Lim as n→∞ Σ(f(n)*Δx)=∫f(x)
Closed integrals
If there is no interval, there is no integral
∫[a,a] f(x)dx=0
Reverse integrals
When the integral is flipped, it becomes negative
∫[a,b] f(x)dx= -∫[b,a] f(x)dx=0
Trapezoid area (under a linear function)
On the interval [a,b]
A(x)=Δx*[f(a)+f(b)]/2
Fundamental theorem of Calculus (Part 1)
Memorize word for word
“If f is continuous on [a,b], then the area function A(x)=∫[a,x] f(t)dt.
For a<b></b>
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Fundamental theorem of Calculus (Part 2)
Memorize word for word
“If f is continuous on [a,b] and F is any antiderivative of f, then
∫[a,b] f(x)dx =F(b)-F(a)”
Integral of an odd function
If on the interval [-a,a]
Then always equal to 0
Integral of an even function
If on the interval [-a,a],
then 2*([0,a] f(x)dx)
Average function value from integrals
(∫[a,b]f(x)dx)/(b-a)
Substituting within integrals
When you have f(g(x)) you must also put g'(x)dx Substitute u=g(x) du=g'(x)dx f(u)du The interval [a,b] becomes [g(a),g(b)]
Distance Traveled
s(b)-s(a)= |∫v(t)| = |∫∫a(t)|
On interval [a,b]
Position from velocity
s(t)=s(ti)+∫v(x)dx
interval [ti,t]
Velocity from Acceleration
v(t)=v(ti)+∫a(x)dx
interval [ti,t]
‘Net’ values
f(b)-f(a)= ∫f’(t)dt
On interval [a,b]
Future values of a function
f(t)=f(ti)+∫f’(x)dx
interval [ti,t]
Region between two curves (from x)
A= ∫[f(x)-g(x)]dx
interval [a,b]
Region between two curves (from y)
A= ∫[f(y)-g(y)]dy
interval [c,d]
When x=f(y) and x=g(y)
Needs practice
Volume by slicing
When b-a equals the object width (cross section on the x-axis)
And A(x) is the ‘area under the curve’ on a z-plane at the x-value
V= ∫A(x)dx
interval [a,b]
Total Volume by washer method (axis-rotation arround x)
When b-a equals the object net width (cross section on the x-axis)
f(x) describes the outer ‘hight’ from the cross-section cut (outer radius)
g(x) describes the inner ‘hight’ from the cross-section cut (inner radius
Where f(x)-g(x) is the net height at x
V= ∫π*(f(x)^2-g(x)^2)dx
interval [a,b]
Total Volume by washer method (axis-rotation arround y)
When b-a equals net object height (cross section on the y-axis)
f(y) describes the outer ‘width’ from the cross-section cut (outer radius)
g(y) describes the inner ‘width’ from the cross-section cut (inner radius)
V= ∫π*(f(y)^2-g(y)^2)dy
interval [a,b]
Total Volume by shell method (axis-rotation arround y)
When b-a equals half the object net width (cross section on the y-axis)
f(x) describes the ‘height’ of the top of the object from the x-axis (x=inner radius at that f(x)
g(x) describes the ‘height’ of the bottom of the object from the x-axis (x=outer radius at that g(x)
Where f(x)-g(x) is the actual object net height at x
V= ∫2πx*(f(x)-g(x))dx
interval [a,b]
Total Volume by shell method (axis-rotation arround x)
When b-a equals half the object net height (cross section on the x-axis)
f(y) describes the rightmost radius from the x-axis
g(y) describes the leftmost radius from the x-axis
V= ∫2πy*(f(y)-g(y))dy
interval [a,b]
Arc length on an interval [a,b]
Arc= ∫√(1+f’(x)^2)dx
Hooke’s Law (Spring Force)
F(x)=kx
K- spring coefficient, measure of the stiffness of that spring
X- distance pulled
Work required to lift inside a container
W= ∫DGA(y)H(y) dy
On the interval [a,b]
D- material desity
G- gravity contant: 9.8 on earth, 0 in space
A(y)- Cross-sectional area as a function of height ‘y’ (usually πr^2)
H(y)- Distance the water is lifted as a function of height ‘y’, represents net liquid height
Pressure
P(y)=(DA(y)H(y)*G)/S(y)
D- material desity
G- gravity contant: 9.8 on earth, 0 in space
A(y)- Cross-sectional area as a function of height ‘y’ (usually π*r^2)
S(y)- surface area as a function of height’y’
H(y)- Distance the water is lifted as a function of height ‘y’, represents net liquid height
Force of materials
F=∫DG(a-y)*w(y) dy
On the interval [0,a]
D- material desity
G- gravity contant: 9.8 on earth, 0 in space
(a-y)- net depth
w(y)- surface width as a function of height ‘y’
Integral ∫1/x dx
Ln x
Simple as that
Integral ∫ln x^p
∫ln x^p= p*(∫ln x)
Integral ∫ln(x/y)
∫ln(x/y)= ∫ln(x)-∫ln(y)
Integral ∫ln(xy)
∫ln(xy)= ∫ln(x)+∫ln(y)
Integral ∫[f’(x)/f(x)]
∫[f’(x)/f(x)]= ln |f(x)|
Integral ∫e^x
∫e^x= e^x
Integral ∫b^x dx
∫b^x dx =(b^x)/(ln b) +C
Integration by parts
∫f(x)g’(x)dx= f(x)*g(x)-∫g(x)f’(x)dx
Integral ∫ln(x)
∫ln(x)dx=x*ln(x)-x+C
Integral ∫sin^n(x)dx
∫sin^n(x)dx =[-sin^(n-1)xcos(x)]/n+(n-1)/n∫sin^(n-2)x dx
Write it out
Integral ∫cos^n(x)dx
∫cos^n(x)dx =[cos^(n-1)xsin(x)]/n+(n-1)/n∫cos^(n-2)x dx
Write it out
Integral ∫tan^n(x)dx
∫tan^n(x)dx= [tan^(n-1)x]/(n-1)-∫tan^(n-2)x dx
∫sec^n(x)dx
∫sec^n(x)dx =[sec^(n-2)xtan(x)]/(n-1)+(n-2)/(n-1)∫sec^(n-2)x dx
Write it out
Integral ∫tan(x)dx
∫tan(x)dx =-ln|cos(x)|+C =ln|sec(x)|+C
Integral ∫cot(x)dx
∫cot(x)dx =ln|sin(x)|+C
Integral ∫sec(x)dx
∫sec(x)dx =ln|sec(x)+tan(x)|+C
Integral ∫csc(x)dx
∫csc(x)dx =-ln|csc(x)+cot(x)|+C
Partial fraction decomposition (before integrating rational functions)
1) Completely factor denominator
2) Set original function equal to sum of denominator factors with alphabetical (A, B, C, …) numerators
3) work the algebra
Absolute Error
(ActualIntegral-RiemannSum)
Relative error
(ActualIntegral-RiemannSum)/ActualIntegral
Trapezoid Rule
Σ ([f(x-1)-f(x)]/2)Δx
Integral Symmetry
D
Integrals with Buoyance
D
Archimede’s Principal
D
Integrals Hyperbolic functions
D
Euler’s method in integral
D
Simpson’s Rule
F
Preditor-prey integration models
G
Mercator Projections
D
Integrals in logistic growth
D
Work
The use of energy to create a force
Given in NewtonMeters, the product of force and displacement or the integral of Force as a function of displacement when graphed against eachother
W=F*Δs=∫F(Δs)
Antiderivatives F(x)
Basically the integral…
The antiderivative is the function to that derivative-function,
F’(x)=f(x)
Position is the antiderivative to a velocity curve
