INTERGRAL CALCULUS Flashcards
a branch of mathematics concerned with the theory and applications (as in the determination of lengths, areas, and volumes and in the solution of differential equations) of integrals and integration
integral calculus
result of the integration is called
the integral
two types of integral
definite and indefinite
before integration, the function to be integrated is called
integrand
integral of the ff: sin x = cos x = tan x = cot x = csc x = sec x =
sin x = -cos x + C cos x = sin x + C tan x = -ln |cos x| + C cot x = ln |sin x| + C csc x = -ln |csc x + cot x| + C sec x = ln |sec x + tan x| + C
integral of the ff:
∫ log(a) xdx = ?
∫ ln xdx = ?
∫ log(a) xdx = x ( log(a) x - log(a) e ) + C
∫ ln xdx = x ln(x) -x +C
∫ sec^2 xdx= ∫ csc^2 xdx= ∫ sec x tan xdx= ∫ csc x cot xdx= ∫ 1/(x^2 +1) dx = ∫ sinh xdx = ∫ cosh xdx = ∫ 1/x dx= ∫ b^x dx= ∫ 1/√(1-x^2) dx=
∫ sec^2 xdx= tan x + C ∫ csc^2 xdx= -cot x + C ∫ sec x tan xdx= sec x + C ∫ csc x cot xdx= -csc x + C ∫ 1/(x^2 +1) dx = tan^-1 x + C ∫ sinh xdx = cosh x + C ∫ cosh xdx = sinh x +C ∫ 1/x dx= ln |x| + C ∫ b^x dx= b^x / ln(b) + C ∫ 1/√(1-x^2) dx= sin^-1 x +C
also known as u-substitution or change of variables, is a method for evaluating integrals and antiderivatives.
subtitution rule
given the ff expression, give their respective trigonometric subtitution for integration and the right identity to be applied
√(a^2 - x^2) , subtitution=? identity=?
√(a^2 + x^2), subtitution=? identity=?
√(x^2 - a^2), subtitution=? identity=?
√(a^2 - x^2) ,
subtitution: , let x = a sin θ
identity: ( 1- sin^2 θ) = cos θ
√(a^2 + x^2),
subtitution: let x = atan θ
identity: 1 + tan^2 θ = sec^2 θ
√(x^2 - a^2),
subtitution: let x = asecθ
identity: sec^2 θ -1 = tan^2 θ
long solution…..
solve for
∫ √(9-x^2)/ x^2 dx
ans : - cot θ - θ + C
step 1 : use trigo sub.
√(a^2 - x^2) ,
subtitution: , let x = a sin θ
identity: ( 1- sin^2 θ) = cos θ
step 2: use to simplify
cots^2 θ + 1 = csc^2θ
step 3: use ;
∫ csc^2 xdx= -cot x + C
step4 : answer is
ans : - cot θ - θ + C
integration by parts formula and
how to know the order of choosing u and dv
∫ vdu = uv- ∫ vdu
LIATE
P= f(x,y)
∂P/∂x means that?
derivative of the function of x and y with respect to x,
x is differntiated and y is constant.
how do you solve a miltiple integration.
solve first the inner integration and move outwards:
- take note of the dx , dy ,dz to know which is your variable and constants.
how do you solve the area under the curve.
give the steps
1 visualize the curve
2 choose a strip (vertical or horizontal)
* strips must touch both lines
3 reqrite equation
* horizontal = dy : stacking multiple strips vertically
* vertical strip= dx ; stacking multiple strips
horizontally
4 determine the limits.
5 solve….
A= ∫(x2-x1) [ Yupper - Ylower]dx
or
A= ∫(y2-y1) [ Xright - Xleft]dy
how to solve if there are three curves involved
just to know the intersection and solv efor the area 2 lines at a time
how to solve if there is multiple area bwet two curves.
just use absolute
A= ∫(x3-x1) | f(x) - g(x)| dx
formula for the area of a sector
A = 1/2 r^2 θ
**a sector is used instead of rectangle when integrating with polar coordinate to find area under a curve.
A = ∫(θ2-θ1) [ 1/2 r^2 θ dθ
when evaluating an area under a curve for POLAR COORDINATES, how do you determing the upper limit and the lower limit?
for rectangular, you use upper- lower & righ - left.
but with polar curves, you do clockwise like (330°-210°)
- use radian for ease
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