Calc 4 Flashcards
LRAM
rectangles left corner touches th ecurve
x: left corner x value
y: height of x value
(sum of heights)(width of each rectangle)= area
MRAM
.5h(LRAM + RRAM)
middle of rectangle touches curve
usually the most accurate
inscribed rectangles
all lie below curve
circumscribed rectangles
all lie above the curve
trapezoidal rule find area
width of subinterval/2 [f(x) + 2f(x1) + 2f(x2) + f(x)}
indefinite integration
antiderivative=intergrand
F'(x)= f(x)
{f(x)dx
-
-add one to exponent
-use scalar rule
-divide by new exponent
-ADD +C
when dividing by negative exponent, whole new terms becomes negative
{1/x dx= ln[x]
find c at a given point
get antiderivative then plug in
reimann sum
find area of each rectangle and add it together. subinterval may be different so if X is larger than one, find MRAM
as width (X) approaches 0, # of rectangles approaches infinity
subinterval
width of a rectangle
longest is [[P]]
partition
entire interval
velocity
area is distance traveled
calculator
math 9
definite integral
if continous
F(b) - F(a)
area: absolute value
displacement: substract
find antiderivative first, then plug a & b into F(X)
definite integral theorem
scalar
sum/diff
average value
is the area of the function is constant over that specific interval.
height of a rectangle that has the same area as under the curve
- find definite integral n divide over the number of values btw a/b
mean value theorem
1/b-a { f(x)dx = f(c )
C is the point when the y value of the curve is same as height of rectangle
fundamental theorem of calculus
taking the derivative of the intergral
b= X a= constant
f’= t instead of x
f(x)= { f’(t)dt
replace t w/ X
if b= u
then it is u’f(u)
net area
finding integral. displacement
to find actual area, divide up the interval n change the negative area to positive n add to positive area
u-sub type 1
define u find u' replace u n du solve for antiderivative plug in real u
u-sub type 2
contant attached to dx, move it over to du n add infront of integral
ISOLATE dx unless its in the integral
u-sub type 3
can’t use product or quotient rule
define u then isolate x n plug in before getting rid of derivatives
u-sub with definite integral
adjust bounds
plug in b/a into u equation to find u= new bounds
use same thing for regular u-sub. bring scalar out
d/dx sin
cos
d/dx cos
-sin
d/dx tan
sec^2
d/dx sec
sectan
d/dx csc
-csccot
d/dx cot
-csc^2
to find position
s(a) + {v(t)dt
s(a)= starting position
to find distance not displacement
figure out when v(t) is 0
divide integral up n solve for antiderivative/position
to find max accel
find a’
then plug into a equation to c which value is higher
integral
1
/ 1- x^2
arcsinx+c
integral
1
1+x^2
arctanx +c
integral
1
[x] /x^2-1
arcsecx
trig substitution
set up triangle
make an equation with x/constant
take derivative of both sides
set @=something
new function that includes rest of integral
subsitite
find antiderivative
if definite then reset bounds w/ @=something
u=asin@
a^-u^
u=atan@
a^+u^
u=asec@
u^-a^
