Integration Flashcards
Sum rule
If f(x) = g(x) + n(x), Then ∫ f(x) dx = ∫ g(x) dx + ∫ n(x) dx
- Integrate term by term
Constants rule
∫ ax^n dx = ax^(n+1) / (n+1)
- Leave constants out of integration, then multiply in at the end
Calculating the constant of integration
- Integrate to obtain the original function
- Substitute in given values from question
- Solve equation to find ‘c’ value
- Rewrite integrated function with ‘c’ value (and the original x/y variables)
- If beginning from the second derivative, integrate twice to obtain the original function. This will produce two constants of integration.
Integrating exponential functions
The answer will be the original function divided by the differentiated value of the power, + c ( x 1/x’)
- Leave constants out of integration, then multiply in at the end
Integrating log functions
- Break down main fraction into smaller fractions
- Apply sum rule
- Integrate each term separately
- Rewrite integrated function, + c
- Terms on the bottom half of the fraction can be moved to the top by inverting the sign of the power
Integrating trigonometric functions
The answer will be the integrated trig function divided by the differentiated value of the ‘x’ value, + c ( x 1/x’)
- Leave constants out of integration, then multiply in at the end
Integrating trigonometric products
- Rewrite function to be integrated as sums (using trig product rules)
- Apply sum rule
- Integrate each term separately
- Simplify and rewrite integrated function, + c
- Leave constants out of integration, then multiply in at the end
Guess/check method for integrating products
If question is f [g(x)]^n
- Try differentiating g(x)^(n+1)
If question is f (e)^g(x)
- Try differentiating (e)^g(x)
If question is f [trig(g(x))]
- Try differentiating ‘differentiation result’ of trig (g(x))
Then from the result of the guess, determine the factor needed to reach the same guess value tried. Then, add this factor to the guess value tried and rewrite function, + c
Guess/check method for integrating quotients
If question is f’(x)/f(x)
- Try differentiating f(x)
Then from the result of the guess, determine the factor needed to reach the same guess value tried. Then, add this factor to the guess value tried and rewrite function, + c
Integrating rational functions in the form (ax+b) / (cx + d)
Divide the denominator into the numerator, then integrate
Integrating by substitution
- Write the expression to be integrated in terms of u instead of x
- Adjust for integrating with respect to u instead of x.
- u = term
- du/dx = differentiated term
- du = differentiated term (dx)
- dx = (du)/differentiated term
- In the integral, after making the substitution with u, replace dx with its equivalent value
eg. (du)/differentiated term - After completing the integration ‘substitute back’, so the final answer is in terms of x, not u.
Properties of definite integrals - Sign
- b∫a f(x) dx = -a∫b f(x) dx
- Width of each integral is now (b-a)/n, hence the base of each rectangle is negative. If the order of the limits of integration are reversed, then the sign of the integral changes.
Properties of definite integrals - Width = 0
- a∫a f(x) dx = 0
- Width of each integral is 0.
Properties of definite integrals - Additive
- b∫a f(x) dx + c∫b f(x) dx = c∫a f(x) dx
- Area is additive
Properties of even functions
- A function f(x) is even if f(-x) = f(x) for all values of x
- Function has the y-axis as an axis of symmetry
Properties of odd functions
- A function f(x) is odd if f(-x) = -f(x) for all values of x
- Function has point symmetry (half-turn rotational symmetry) about the origin
Properties of periodic functions
- A function f(x) is periodic if f(x) = f(x+a) for all values of x and some fixed, non-zero value ‘a’.
- Function repeats itself at regular intervals
- Smallest positive value of the fixed number ‘a’ is the period of the function.
Integration properties of even functions
If a∫-a f(x) dx = 2 a∫0 f(x) dx,
Then function is even
Integration properties of odd functions
If a∫-a f(x) dx = 0,
Then function is odd
Areas with the x-axis as a boundary
b∫a f(x) dx gives the area between f(x) and the x-axis, bounded on the left by a vertical line x = a, and bounded on the right by the vertical line x = b (b is larger value).
Signed areas (areas below the x-axis)
b∫a f(x) dx gives the area between f(x) and the x-axis, bounded on the left by a vertical line x = a, and bounded on the right by the vertical line x = b (b is larger value).
- Result will be negative, but give positive final answer as area cannot be negative
Areas above and below the x-axis
Calculate the areas above and below the x-axis separately, then add them
- Results for areas below the x-axis will be negative, but give positive final answer as area cannot be negative
Areas between two curves
b∫a [f(x) - g(x)] dx gives the area enclosed by two curves f(x) and g(x), and the lines x = a and x = b, where the function f(x) is above g(x).
Areas with the y-axis as a boundary
- Rewrite the curve as a function of y (x = ___)
2. Integrate with respect to y
- Trapezium rule
The approximated area (T) is given by T = h/2 [y0 + 2y1 + 2y2 + … + 2yn-1 + yn]
Where,
h is the interval length (different between successive x values)
n is number of trapezia
Using trapezium/simpsons rules if not given x/y values
Evaluate function values of x between the limits of integration in steps of 0.1
- Simpsons rule
The approximated area (S) is given by S = h/3 [y0 + 4y1 + 2y2 + 4y3 + 2y4 + 4y5 + … + 2yn-2 + 4yn-1 + yn]
Where,
h is the interval length (different between successive x values)
n+1 is the number of y-values
(number of y-value must be odd, therefore the number of different intervals must be even)
Differential equation for when the rate of change is proportional to x
dy/dx = kx
Differential equation for when the rate of change is inversely proportional to x
dy/dx = k/x
Differential equation for when the rate of change is proportional to the amount itself
dy/dx = ky
Checking solutions of differential equations
- Calculate the derivative(s) of the function by integrating both sides
- Substitute the function and its derivative(s) into the differential equation to check that it is true
Solving differential equations in the form dy/dx = f(x)
- Use integration on both sides
- If the equation is a simple second-order differential equation, integrate both sides twice. This gives a general solution that has two constants.
Solving first-order differential equations in the form dy/dx = f(x) x g(x)
- Separate the variables by writing the equation with the x-terms on one side and the y-terms on the other
- Use integration on both sides
The constant in differential equations
- Only one constant turns up in the solution of a first-order differential equation
- When integrating expressions that have a log function as their anti-derivative (eg. 1/x), the constant of integration can be a multiplicative one (k) inside the logarithm function rather than additive (+c)
Obtaining particular solutions from differential equations when given boundary conditions
- Solve the differential equation
- Substitute x and y values from equation into the equation and solve to find ‘c’
- Rewrite equation with ‘c’ value (and the original x/y variables)
Applications of differential equations
- Solve the equation, introducing a constant as part of the solution
- Use the information given to evaluate the constant and substitute back to obtain a particular solution
- Use the particular solution to answer some question
Application differential equation for when a quantity is changing at a steady rate
dy/dt = k,
y = kt + c
Application differential equation for when the rate of change of a quantity is some function of time
dy/dt = f(t),
y = ∫ f(t) dt
Application differential equation for when the rate of change of a quantity is proportional to the quantity itself
dy/dt = ky,
y = Ae^kt
Application for when the differential equation has separate variables
Separate the variables and then integrate
Newtons law of cooling
dT/dt = -k(T - To)
Where,
T is the temperature of the hot body
To is the temperature of its surroundings
Application of Newtons law of cooling
T = To + Ke^-kt
- Two conditions (T and t) are required to solve for the two constants involved (k and K)
Integral of tan(x)
-ln |cos(x)|
Integral of 1/tan(x)
ln |sin(x)|
Equivalent of tan(x)
sin(x)/cos(x)
