linear functions - mrs. young's math...

of Area under the Curve

To approximate the area between a function f (x) and the x-axis, create rectangles that best fit and find the sum of the area of these rectangles. A curve above the x-axis is said to enclose “positive” area while a curve below the x-axis encloses “negative” area.

Integral 1 of 24

of

Riemann Sum: LRAM, MRAM, RRAM

Riemann Sum uses regular intervals when creating rectangles to approximate the area under the curve, Rectangular Approximation Method (RAM). LRAM creates rectangles with the height determined by the upper left-hand corner intersecting with

Integral 2 of 24

of the function. In MRAM, the function intersects at the upper middle part of the rectangle. In RRAM, the height of each rectangle is determined by the upper right-hand corner. LRAM, MRAM, and RRAM can be seen in that respective order below.

Sn=∑k=1

n

f (ck) Δ xn=number of rectanglesΔ x=widthof rectangles f ( ck )=height of eachrectnalge

Trapezoidal Rule

Trapezoidal rule uses trapezoids to approximate the area between a function f (x) and the x-axis. The height, h, of each trapezoid is the constant interval Δ x and the bases are the y values at each interval.

Integral 3 of 24

of

T=h2( y0+2 y1+2 y2+…+2 yn−1+ yn)

Definite Integral

Integral 4 of 24

of

As the width of each rectangle Δ x decreases, the accuracy of the area approximation increases. Creating more rectangles over an interval [a ,b] yields a better answer. Thus, breaking an interval into infinitely lim n→ ∞ many rectangles yields an integral from a to b. dx keeps track of variable of integration and comparable to ∆ x .

∫a

b

f ( x )dx= limn→ ∞

∑k =1

n

f (ck) Δ x

Indefinite Integrals

Integral 5 of 24

of Antidifferentiation is the inverse of differentiation: an antiderivative of f (x) is any function F (x) whose derivative is equal to f (x): F ' ( x )=f (x). There are many functions with the same derivative; these functions usually differ by a constant C. Therefore, the family of antiderivatives of f (x) are F ( x )+C. Thus, the indefinite integral of f (x):

∫ f ( x )dx=F ( x )+C

∫ ( 3 x2+4 x−8 ) dx=x3+2 x2−8 x+C

∫ csc x cot xdx=−csc x+C

∫ 1x2+1

dx=tan−1 x+C

∫ dxx

=ln|x|+C

Fundamental Theorem of CalculusIntegral 6 of 24

of Part 1

If f is continuous on [a , b], then the function

F ( x )=∫a

x

f (t ) dt

Has a derivative at every point x in [a , b], and

F ' ( x )= ddx∫a

x

f ( t ) dt=f (x)

Part 2 (Integral Evaluation Theorem)

If f is continuous at every point of [a ,b], and if F is an antiderivative of f on [a ,b], then

∫a

b

f ( x )dx=F (b )−F (a)

Integral 7 of 24

of Integral Properties

f (x) and g(x ) are separate functions

∫ ( f ( x )± g ( x ) ) dx=∫ f ( x )dx ±∫ g ( x ) dx

Constant Multiple: k is any real number

∫ kf (x ) dx=k∫ f ( x ) dx

Additivity: for any b, such that a<b<c

∫a

b

f ( x )dx+∫b

c

f ( x ) dx=∫a

c

f ( x ) dx

Order of Integration

∫a

b

f ( x )dx=−∫b

a

f ( x )dx

Zero Integral

∫a

a

f ( x )dx=0

Domination: f (x)≥ g (x)

∫a

b

f ( x )dx ≥∫a

b

g ( x ) dx

Integral 8 of 24

of

∫ dx|x|√x2−1

=sec−1 x+C

u-Substitution

This is the chain rule for integrals and it is used for composite functions f (g(x )) and products. Let u=g (x), then

∫ f ( g (x ) ) g' (x ) dx=∫ f (u)du

or, with limits,

∫a

b

f ( g (x ) ) g' (x ) dx=∫g(a)

g(b)

f (u)du

Example:

∫ x2 √x3+8dx

Integral 10 of 24

of

u=x3+8 du=3 x3 dx⟹ du3

=x2 dx

∫ x2 √x3+8dx=13∫ √udu=1

3∙ 23

u32+C=2

9 √(x3+8)3+C

Piecewise Integration

h ( x )={f (x ) x≥ bg(x ) x<b

If a<b<c , then the integral of h(x ) from a to c is

∫a

c

h ( x ) dx=∫a

b

f ( x ) dx+∫b

c

g (x ) dx

Absolute Value IntegrationRewrite as piecewise function. Solve for x inside the absolute value symbol. The inside of the absolute value should be positive with x ≥ and negative with x<¿.

Integral 11 of 24

of Then, integrate as a piecewise function.

|f (x)|={ f ( x) x ≥b−f (x) x<b

If a<b<c , then the integral of f (x) from a to c is

∫a

c

|f ( x )|dx=∫a

b

f ( x ) dx−∫b

c

f ( x ) dx

Original Equation

Given a derivative and an initial value, to find the original equation:

1. Take the antiderivative.2. Substitute the initial value.3. Solve for the constant, C.4. Rewrite equation.

Examples

Integral 12 of 24

of Find the particular solution to the

equation dydx

=ex−6 x2 whose graph

passes through the point (1, 0).

y=∫ ( ex−6 x2 ) dx=ex−2 x3+CApply the initial condition (1, 0).

0=e1−2(1)3+CSolve for C.

C=2−eRewrite Equation.

y=ex−2 x3+2−e

Find the particular solution to the

equation dydx

=2x−sec2 x whose

graph passes through the point (0, 3).

y=∫ ( 2 x−sec2 x ) dx=x2− tan x+CApply the initial condition (0, 3).

3=(0)2−tan (0)+CSolve for C.

C=3Rewrite Equation.

y=x2−tan x+3Physical Motion

Displacement: s( t )

Velocity: v(t )=s' (t )

Acceleration: a ( t )=v ' ( t )=s' ' (t)

Conversely,

v ( t )=v0+∫0

t

a ( x )dx

Integral 13 of 24

of

s ( t )=s0+∫0

t

v ( x ) dx

Total distance traveled from t=a to t=b is

∫a

b

|v (t)|dt

Example

A particle travels with velocity v ( t )=(t−2 ) sin t m/sec for 0 ≤ t ≤ 4 sec. What is the particle’s displacement? What is the total distance traveled?

displacement=∫0

4

(t−2 ) sin t dt ≈−1.45 meters

distance traveled=∫0

4

|( t−2 ) sin t|dt ≈ 1.91 meters

Consumption over Time

Integral 14 of 24

of Let C (t) be a rate at which something in consumed over a period from t=a to t=b . Then, the total consumption can be modeled by the integral

T c=∫a

b

C ( t ) dt

Hooke’s Law

Hooke’s Law for springs says that the force it takes to stretch or compress a spring x units from its natural (unstressed) length is a constant times x: F=kx . The amount of Work needed to stretch or compress a spring x units from its natural (unstressed) length can be found by solving the integral below.

∫0

b

F (x ) dx=∫0

b

kxdx

Area

Integral 15 of 24

of If f and g are continuous with f (x)≥ g (x) throughout [a , b], then the area between the curves from a to b is the integral of [ f −g ] from a to b,

A=∫a

b

[ f (x )−g(x )] dx

Example

Find the area of the region enclosed by the parabola y=2−x2 and the line y=−x.

The limit of integration are found by finding where the two equations intersect 2−x2=−x⟹ x=−1 ,2.

A=∫−1

2

[ (2−x2 )−(−x )] dx=92

Integral 16 of 24

of Thus, the area of the region is 4.5 square units.

Volume

If a solid in space is oriented so that the area of a cross-section perpendicular to the x-axis is given by A(x ). Then, the volume of solid bounded by the

planes x=a and x=b is

∫a

b

A (x)dx

Disk MethodVolume of a solid that revolves around the x-axis bounded by x=a and x=b is

Integral 17 of 24

of

∫a

b

π (radius )2 dx=π∫a

b

( f ( x ))2dx

Shell MethodVolume of a solid that revolves around the y-axis bounded by x=a and x=b is

∫a

b

2 π ( shellradius)( shell

height )dx

¿2 π∫a

b

xf (x )dx

Washer Method

Integral 18 of 24

of If f (x)≥ g (x) between a and b, the volume of a solid that revolves around the x-axis bounded by x=a and x=b is

∫a

b

π ( outerradius)

2

−π ( innerradius)

2

dx

¿ π∫a

b

[( f ( x ))2−(g ( x ))2 ]dx

Mean (Average) Value Theorem for Integrals

If f is continuous on [a ,b], then at some point c in [a , b],

Integral 19 of 24

of

f ( c )= 1b−a∫a

b

f ( x ) dx

The Mean Value Theorem for Integrals is the average value of a function over an interval. This is not to be confused by the Mean Value Theorem for Derivatives. The MVT for Derivatives is the average rate of change of a function over an interval.

Example

What is the average value of the cosine function on the interval [1,5]?

f ( c )= 15−1∫1

5

cos x dx=−0.450

The average value of cosine is –0.450.

If the average value of the function f on

the interval [ a ,b ] is 10, then ∫a

b

f ( x )dx?

10= 1b−a∫a

b

f ( x ) dx

∫a

b

f ( x )dx is equal to 10(b−a).

Integral 20 of 24

of Separable Differential Equations

A differential equation expressed in the form dydx

=f (x)g ( y ) is called separable, if

it is possible to separate the variables completely. To integrate separable differential, separate the variables and rewrite in the form

1g ( y )

dy=f (x )dx

Then integrate each side separately. It is sufficient to write only one constant, C.

Exampledydx

=6 x2 y2⇒ dyy2 =6x2 dx

∫ dyy2 =∫6 x2 dx

−1y

=2 x3+C

y= −12 x3+C

dydx

=( y+2 ) (2x−3 )⇒ dyy+2

=(2x−3 ) dx

∫ dyy+2

=∫(2 x−3)dx

Integral 21 of 24

of ln ( y+2)=x2−3x+C

y+2=Ce x2−3 x

y=Cex2−3 x−2

Exponential Growth and Decay

Law of Exponential Change

If y changes at a rate proportional to the amount present ( dydt

=ky ) and y0 is the

initial amount at t=0, then y= y0ekt. The constant k is the growth constant if k>0 or the decay constant if k<0.

Half-Life

The half-life, h, of a radioactive substance with rate constant k is h= ln2k

.

Modeling Growth with other Bases

y= y0bht where the base b is the multiplied by y0 over the time period 1h

.

Integral 22 of 24

of Example

Scientists who use carbon-14 dating use 5700 years for its half-life. Find the age of a sample in which 10% of the radioactive nuclei originally present have decayed.

y= y0bht⇒ 0.9 y0= y0( 12 )

t5700⇒ t=5700( ln0.9

ln0.5 )=866

The sample is about 866 years old.Slope Fields

A slope field graphically shows the family of solution to a differentiable equation in dydx

=f (x)g ( y ) form. To construct a slope field, for each point (x , y ), draw a

short line segment with slow dydx

. To interpret a slope field (or find a particular

solution) at a given point, follow the shape of the slope field through that point.

Integral 23 of 24

of

(a) A slope field for the differential equation dydx

=x+ y

(b) The same slope field with the graph of the solution through (2,0) superimposed.

Integral 24 of 24

linear functions - mrs. young's math...

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