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MATH 116 Final Exam-AID Session Zahra Mahmood BodlaPowerPoint Presentation

MATH 116 Final Exam-AID Session Zahra Mahmood Bodla

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MATH 116 Final Exam-AID Session Zahra Mahmood Bodla

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MATH 116 Final Exam-AID SessionZahra MahmoodBodla

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A function is a rule that associates exactly one output value to a given input value.

Notation: y = f(x)

Domain: set of allowable input values.

Range: set of all possible output values.

The Vertical Line Test: A curve is a function in the x-y plane iff the curve does not intersect a vertical line more than once.

Examples

- y = x2 + 1
Domain: ℝ

Range: (1, ∞)

- x2 + y2 = 9 is not a function as it does not pass the vertical line test

A function f is even if the graph of f is symmetric with respect to the y axis; algebraically f(-x) = f(x).

A function f is odd if the graph of f is symmetric with respect to the origon; algebraically f(-x) = -f(x).

A periodic function is a function that repeats itself after some given period, or cycle; mathematically f(t) = f(t + nT), where n is an integer and T is the period.

Examples

A function f(x) has graph:

Even periodic extension of f(x):

Odd periodic extension of f(x):

Determine if following functions are even or odd or neither

a) f(x)=sqrt(x)

b) f(x)=x*abs(x)

Solutions:

Substitute –x as input to each function

a) f(-x)=sqrt(-x), we cannot perform any algebraic operation henceforth, this function is neither odd or even.

b) f(-x)=( -x)*abs(-x)

f(-x)=-x*abs(x), f(-x)=-f(x), therefore odd

Absolute Value Function: a function that gives the magnitude of its input values, example absolute value of -3 is 3.

Composite Functions

Let g(x) have domain D1 and Range R1 , and f(x) have domain D2 ⊃ R1 then the composition of f and g is the function f o g defined by; (f o g)(x) = f(g(x))

Example:

If g(x)=x2-1, and f(g(x))=sqrt(g(x)), then the domain of f(g(x)) is such that g(x)>=0, which is iff x<=-1 or x>=1. So the domain of f(g(x)) is therefore (-∞,-1]U[1,∞), and the range is [0,∞).

A function is called one to one for any x1 , x2 in the domain of f with x1 not equal to x2the f(x1) is not equal to f(x2).

y=x2, not one-to-one

y=x3, one-to-one

The inverse of a function: If f is one-to-one with domain A and range B. Then its inverse f-1, is defined as

f-1(y)=x ifff(x)=y, with domain B and range A.

Basically an inverse of function takes the output of f and returns the corresponding input.

When finding the inverse of a function:

- Check if the function is one-to-one
- Solve the equation for x in terms of y, f(x)
- Then interchange them to get f-1(x)
Inverse Property: If f(x) and g(x) are inverses of each other, then f(g(x))=x and g(f(x))=x.

i.e.

if f(x)=x^3-1, g(x)=(x+1)^(1/3), f and g are inverses of each other.

f(g(x))=((x+1)^(1/3))^3-1

= x+1-1

=x

Conversely you can check g(f(x))=x

Trigonometric functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications. The most basic and important trig. Functions are: sin(x), cos(x) and tan(x).

Inverse trigonometric functions are the inverse functions of the trigonometric functions, though they do not meet the official definition for inverse functions as their ranges are subsets of the domains of the original functions. Since none of the trigonometric functions are one-to-one, they must be restricted in order to have inverse functions.

For example, just as the square root function is defined such that y2 = x, the function y = arcsin(x) is defined so that sin(y) = x. There are multiple numbers y such that sin(y) = x; for example, sin(0) = 0, but also sin(π) = 0, sin(2π) = 0, etc. It follows that the arcsine function is multivalued: arcsin(0) = 0, but also arcsin(0) = π, arcsin(0) = 2π, etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each x in the domain the expression arcsin(x) will evaluate only to a single value, called its principal value. These properties apply to all the inverse trigonometric functions.

Arcsin x

Arctan x

Arccos x

Examples

Evaluate each of the following

we use the following restrictions on inverse cosine :

The restriction on the guarantees that we will only get a single value angle and since we can’t get values of x out of cosine that are larger than 1 or smaller than -1 we also can’t plug these values into an inverse trig function.

So, using these restrictions on the solution we can see that the answer in this case is

The second solution then follows as

Note:

Exponential Functions: functions in the form of ax , where “a” is a constant, example f(x)=3x.

They always have the property with various constants, such that the domain is (-∞,∞) and the range is (0, ∞).

Logarithmic Functions: are the inverse of exponential functions, in the form loga(x), where a is the base constant, example f(x)=log3(x). Observe from the above definition that the domain of various bases of these functions is (0,∞) and the range is (-∞,∞).

ex: is an unique exponential function, such that the slope of tangent line at point x=0 is equal to 1. e is approximately 2.718 correct to three decimal places.

Rules for exponentials and logarithms:

ln(a)+ln(b)=ln(ab)a-n=1/an

ln(a)-ln(b)=ln(a/b) a0=1

ln(an) =nln(a)loga(a)=1

Hyperbolic Functions: combinations of exand e-xarise so frequently in nature, that they are given specific names. These are functions that have the same relationship to a hyperbola as trigonometric functions have a relationship to a circle.

Definitions:

sinh(x)=(ex-e-x)/2 cosh(x)=(ex+e-x)/2

tanh(x)=sinh(x)/cosh(x)

Like trigonometric, there are very useful identities we can use in solving problems involving these functions, which you can check using the function definitions.

sinh(-x)=-sinh(x)

cosh(-x)=cosh(x)

cosh2(x)-sinh2(x)=1

1-tanh2(x)=sech2(x), divide 3. By cosh2(x)

sinh(x+y)=sinh(x)cosh(y)+cosh(x)sinh(y)

cosh(x+y)=cosh(x)cosh(y)+sinh(x)sinh(y)

A limitof a function is used to describe the value a function approaches as input approaches some value.

In general, we say f(x) has limit L as x approaches a if f(x) can be made arbitrarily close to L by taking x sufficiently close to a limx->af(x)=L

Fact: For any polynomiallimx->af(x)=f(a)

Limit Laws

If limx->af(x)=F and limx->ag(x)=G

- limx->a (f(x)+g(x))=F+G
- limx->a (f(x)-g(x))=F-G
- limx->a (f(x).g(x))=FG
- limx->a (f(x)/g(x))=F/G if G is not equal 0

One sided Limits

Left hand limit: limx->a- f(x)=L means f(x) has limit L as x approaches a from the left.

Right hand limit: limx->a+ f(x)=L means f(x) has limit L as x approaches a from the right.

limx->a f(x)=L ifflimx->a- f(x)=L and limx->a+ f(x)=L.

Infinite Limits

We can have limits that approach ∞ or -∞, because as they approach x=a, the output gets infinitely large (e.g. limx->0(1/x2)= ∞), these are called vertical asymptotes.

We can also evaluate the limit of functions approaching infinity, these limits are called horizontal asymptotes (e.g. limx->-∞(1/x2)=0 and limx->+∞(1/x2)=0, x2 greatly dominates 1 as x->∞).

Squeeze Theorem:if f(x)≤g(x)≤h(x), and limf(x)=limh(x)=L, then limg(x)=L, when x→a.

Examples

Find limit of sqrt(x2+1) –x as x→∞

What’s the horizontal asymptote of sqrt(x2+1) - x from [0,∞)?

Since the limit of sqrt(x2+1) –x as x→∞ was 0, we know the horizontal asymptote is x axis.

Find limx→0 (x2/3*cos(1/x2)).

-1≤cos(a)≤1 for any a

-1≤cos(1/x2)≤1

-x2/3≤x2/3cos(1/x2)≤x2/3

limx→0(-x-2/3)= limx→0(x-2/3)=0

Therefore by squeeze theorem limx→0(x2/3cos(1/x2))=0

Continuity

A function f(x) is continuous at x=a if it satisfies

limx->af(x)=f(a) i.e

- limx->af(x) exists
- f is defined at x=a
- limit is equal to the value of the function
A function that does not satisfy one or more of these points is discontinuous at x=a

Types of discontinuities

Infinite (asymptote), jump, hole

Composition Rule for Limits

If limx->ag(x)=L and f(y) is continuous at y=L then limx->a(fog)(x)= limx->a(f(g(x))= f(limx->ag(x))=f(L)

Examples

Find limit of

- limx->0 (sin(x)+cos(x))
- limx->1(x2-1)/(x-1)
- sin(x)+cos(x) is continuous everywhere, so the limit is sin(0)+cos(0)=0+1=1
- We cannot substitute x=1, because the function is not continuous at 1
We have,

limx->1(x2-1)/(x-1)

=limx->1((x+1)(x-1))/(x-1) ``divide out x-1``

=limx->1(x+1), x+1 is continuous at x=1

=1+1

=2

Heavyside Function

The Heavyside function is defined as,

Heaviside functions are often called step functions. Here is some alternate notation for Heaviside functions.

We can think of the Heaviside function as a switch that is off until t = c at which point it turns on and takes a value of 1. So what if we want a switch that will turn on and takes some other value, say 4, or -7?

Heaviside functions can only take values of 0 or 1, but we can use them to get other kinds of switches. For instance 4uc(t) is a switch that is off until t = c and then turns on and takes a value of 4. Likewise, -7uc(t) will be a switch that will take a value of -7 when it turns on.

Derivatives:

The derivative of f(x) is defined by the function f`(x), which is defined as the limit

limx->a(f(x)-f(a))/(x-a)

if we let h=x-a, then we get this limit in a different form, but expressing the same thing and sometimes easier to use.

f`(x)=Limh->0(f(x+h)-f(x))/h

The derivative represents an infinitesimal change in the function with respect to x in this context. Hence the derivative is a function that can be used evaluate the instantaneous rate of change at any point x on the function f(x). (e.g. the derivative of x2 is 2x, which can be obtained using the definition).

Derivatives are particular important motion, the derivative of the position of an object gives its velocity, and the derivative of its velocity gives its acceleration.

Examples

Get derivative of f(x)=abs(x) at x=0 using the definition above.

f’(0)=Limh->0(abs(0+h)-abs(0))/h

f’(0)=Limh->0(abs(h))/h

When dealing with absolute values for input, we have consider when the input is positive and when it is negative

When h<0,abs(h)=(-h)

So Limh->0-(-h/h)

= Limh->0-(-1)

=-1

When h>0,abs(h)=h

So Limh->0+(h/h)

= Limh->0(1)

=1

Since the left limit and right limit are not equal, we know that abs(x) has no derivative at x=0.

Rules for Differentiation

Derivative of a constant function.

The derivative of f(x) = c where c is a constant is given by f '(x) = 0

Derivative of a power function (power rule).

The derivative of f(x) = x r where r is a constant real number is given by f '(x) = r x r - 1Derivative of a function multiplied by a constant.

The derivative of f(x) = c g(x) is given by f '(x) = c g '(x)

Product Rule

Quotient Rule

Chain Rule

Implicit Differentiation: is a method that consists of differentiating both sides of a function and then finding

Examples

Find the derivative of arcsin(x)

Differentiability and Continuity

If f’(a) exists then f(x) is continuous at x=a i.e. if a function is differentiable it is continuous. Corollary: If f(x) is discontinuous at x=a then f’(a) does not exist.

Example

f(x)=|x|

From graph we can see f(x) is continuous everywhere.

However, using the definition of derivative, derivative at x=0

Hence no derivative exists at x=0

Derivatives of Trigonometric and Inverse Trigonometric Functions

Derivatives Of Exponential And Log Functions

Logarithmic Differentiation

The method of logarithmic differentiation ,in calculus, uses the properties of logarithmic functions to differentiate complicated functions and functions where the usual formulas of differentiation do not apply

Examples

y = x sin x

ln y = ln [ x sin x ]

ln y = sin x ln x

y ' / y = cos x ln x + sin x (1/x)

y ' = [ cos x ln x + (1/x) sin x ] y y ' = [ cos x ln x + (1/x) sin x ] x sin x

Derivatives Of Hyperbolic Functions

Rolle’s Theorem: If a function f(x) satisfies

- f(x) is continuous for a≤x≤b
- f’(x) exists for a<x<b
- f(a)=f(b)
then there exists at least one point c with a<c<b such that f’(c)=0

Mean Value Theorem: If a function f(x) satisfies

- f(x) is continuous for a≤x≤b
- f’(x) exists for a<x<b
then there exists at least one point c with a<c<b such that f’(c)=(f(b)-f(a))/(b-a)

Newton’s Method:

The idea of the method is as follows: one starts with an initial guess which is reasonably close to the true root, then the function is approximated by its tangent line (which can be computed using the tools of calculus), and one computes the x-intercept of this tangent line (which is easily done with elementary algebra). This x-intercept will typically be a better approximation to the function's root than the original guess, and the method can be iterated.

If xn is an approximation a solution of f(x)=0 and if f’(xn)≠0 the next approximation is given by,

Increasing and Decreasing Functions

A function f(x) is increasing on an interval I if for all x1> x2 in I, f(x1)> f(x2)

A function f(x) is decreasing on an interval I if for all x1> x2 in I, f(x1)< f(x2)

Increasing/ Decreasing Test

A function is increasing on an interval I if f’(x)≥0 for all x in I and f’(x)=0 at a finite number of points

A function is decreasing on an interval I if f’(x)≦0 for all x in I and f’(x)=0 at a finite number of points

Critical Point: A critical point of a function is a point in the domain of the function where f’(x)=0 or f’(x) does not exist

Relative Maximum: A function has a relative maximum f(x0) at x=x0 if there is an open interval I such that f(x)≤f(x0) for all x in I

Relative Minimum: A function has a relative minimum f(x0) at x=x0 if there is an open interval I such that f(x)≥f(x0) for all x in I

The graph of a function is concave up on an interval I if f’(x) is increasing on I.

The graph of a function is concave down on an interval I if f’(x) is decreasing on I.

Points on the graph of f where concavity changes are called inflection points.

The graph of f(x) is concave up on I, if f’(x)>=0 and f”(x)=0 only at a finite number of points on I.

The graph of f(x) is concave down on I, if f’(x)=<0 and f”(x)=0 only at a finite number of points on I.

Points of inflection occur at points in the domain of f where f”(x) changes sign. [f”(x)=0 or f”(x) does not exist]

Describe the concavity of the graph of

f(x)= 2sin2(x)-x2 for 0≤x≤pi

f’(x)=4sin(x)cos(x)-2x=sin(2x)-2x

f”(x)=4cos(2x)-2

We can find points of inflection by solving 4cos(2x)-2=0 which yields the solution x=pi/6

In fact, we have f”(x)≥0 when 0≤x≤pi/6 and f”(x)≤0 when pi/6≤x≤pi

Therefore, the graph is concave up when 0≤x≤pi/6 and concave down when pi/6≤x≤pi

Suppose f”(x) is continuous on an open interval containing the critical point x0 with f’(x0)=0.

Then

- f”(x)>0 ⇒ f has a local minimum at x=x0
- f”(x)<0 ⇒ f has a local maximum at x=x0
- f”(x)=0 ⇒ no conclusion

Determine the critical points and their nature of f(x)=x4

f’(x)=4x3

f”(x)=12x2

f’(x)=4x3=0 at x=0

f”(x)=12x2 at x=0 f”(x)=0 so we have no conclusion from the second derivative test.

Use first derivative test

f’(x)<0 for x<0 and f’(x)>0 for x>0

So x=0 is a local minimum point

Sketch the graph of f(x)=ex/(2-3ex)

We start by finding the x and y intercepts:

f(x) is never 0 so we have no x intercepts

f(0)=-1 so we have a y intercept at (0,-1)

Next we find the asymptotes:

2-3ex=0 ⇒ Vertical asymptote at x=ln(2/3)

Limx→ln(2/3)+ex/(2-3ex) = -∞ and Limx→ln(2/3)-ex/(2-3ex) = +∞

Horizontal asymptotes:

Limx→ +∞ex/(2-3ex) = -1/3

Limx→ -∞ex/(2-3ex) = 0

We then find the critical points

f’(x)=2ex/(2-3ex)2

f’(x) is never 0 so there are no critical points

f’(x) is not defined at x=ln(2/3) and f’(x)>0 for all x≠ln(2/3) so, f(x) is increasing on [-∞, ln(2/3)], [ln(2/3), ∞]

Lastly we find concavity

f”(x)=2ex(2+3ex)/(2-3ex)3

f”(x) is never 0 and f”(x) is not defined at ln(2/3)

f”(x)>0 if x<ln(2/3) so it is concave up

f”(x)<0 if x>ln(2/3) so it is concave down

The function f(x) has an absolute/ global maximum f(x0) on the interval I if x0 is in I and for all x in I, f(x)≤f(x0).

The function f(x) has an absolute/ global minimumf(x0) on the interval I if x0 is in I and for all x in I, f(x)≥f(x0).

Theorem: If I is a closed interval and f(x) is continuous on I then f(x) has an absolute maximum and an absolute minimum on I.

Theorem: If a function has an absolute minimum/ maximum on I it occurs either at a critical point of f or at an end point f I.

Procedure for finding absolute max and min

- Find critical points of function on interval.
- Evaluate f at critical points and end points of interval.
- Pick largest (absolute max) and smallest (absolute min) values.
Example

Find absolue max and min of f(x)=x2/3-x5/3 on [1/5, 2]

f’(x)=2x-1/3/3-5x2/3/3=0

Critical points x=0 (not in interval) and x=2/5

f(1/5)=0.2736

f(2)=-1.5874 ⇒ absolute min on [1/5, 2]

F(2/5)=0.3258 ⇒ absolute max on [1/5, 2]

L’Hopital’s Rule

If f(x) and g(x) satisfy

- f(x), g(x) are differentiable in an open interval I containing x=a, except possibly at x=a.
- g’(x)≠0 in interval except possibly at x=a
- Limx→af(x)=0=limx→ag(x)
- Limx→af’(x)/g’(x)=L
Then Limx→af(x)/g(x)=L

Examples

Limx→0 sinx/x= Limx→0 (sinx)’/x’= Limx→0cosx/1=1

Limx→12lnx/x-1= Limx→1 (2lnx)’/(x-1)’= Limx→1 2/x=2

Limx→0 ex−1/x2= Limx→0 (ex−1)’/(x2)’=Limx→0 ex/2x= ∞

A function f(x) is called the antiderivative of f(x) on an interval I if F’(x)=f(x) for all x in I.

Theorem: If F(x) is an antiderivative of f(x) on an interval I, then every antiderivative of f(x) on I is of the form F(x)+C, where C is a constant.

The Indefinite Integral of f(x) with respect to x is the set of all possible antiderivatives ∫f(x)dx=F(x)+C

Examples

f(x)=cos(x)

∫cos(x)dx=sin(x)+C

Suppose F(u) is an antiderivative of f(u): F’(u)=f(u) and ∫f(u)du=F(u)+C.

Consider the composition of F(u) and g(x)

Chain Rule: [F(g(x))]’=F’(g(x))g’(x) =f(g(x))g’(x)

Reversing this

∫f(g(x))g’(x)dx=F(g(x))+C

Let u=g(x)

∫f(u)u’dx=F(u)+C=∫f(u)du

Change of Variables formula:

∫f(g(x))g’(x)dx=∫f(u)du where u=g(x)

Evaluate ∫tan(x)dx

∫tan(x)dx=∫sin(x)/cos(x) dx

Let u=cos(x)

du=-sin(x)dx

sin(x)dx=-du

∫(1/cos(x))*sin(x)dx= -∫(1/u)du

=-ln|u|+C

=-ln|cos(x)|+C

=ln|1/cos(x)|+C

Reimann sum of f(x) on a≤x≤b

∑f(xi*) ∆xi where a=x0<x1<…<xn>b is a partition of [a,b]

∆xi =xi-xi-1

xi-1≤xi*≤xi

Size of partition||∆xi||=max|∆xi|

Definite Integral

a∫b f(x)dx=lim||∆xi||→0 ∑f(xi*) ∆xi

Theorem: If f(x) is continuous on a≤x≤b (with a and b finite) then a∫bf(x)dxis defined.

First Fundamental Theorem of Calculus

If f(x) is continuous on a≤x≤b and F(x) is an antiderivative of f(x) on a≤x≤b then

a∫bf(x)dx=F(b)-F(a)=F(x)a|b

The velocity of an object moving on the x-axis is given by v(t)=t2-5t+4 m/s. Find 0∫4 v(t)dt and 0∫4 |v(t)|dt and give a physical representation of each.

0∫4 v(t)dt= 0∫4(t2-5t+4)dt =(t3/3-5t2/2+4t)4|0 =-2m

x(4) is 2m to the left of x(0)

0∫4|v(t)|dt= 0∫1|t2-5t+4|dt +1∫4|-t2+5t-4|dt =(t3/3-5t2/2+4t)1|0 + (-t3/3+5t2/2-4t)4|1

=19/3 m

Total distance travelled between t=0 and t=4

1∫2 sin(ln(x))/x dx

u=lnx

du=dx/x

X=1, u=0

X=2, u=ln2

1∫2 sin(ln(x))/x dx

=0∫ln2 sin(u)du

=-cos(u)0|ln2

=-cos(ln2)-(-cos(0)

=1-cos(ln2)

If f(x) is continuous on a≤x≤bthen the function F(x)=a∫x f(t)dtis differentiable for a≤x≤band F’(x)=f(x) i.e. F(x) is an antiderivative of f(x).

Example

Let h(x)=2∫sinx t2dt, find h’(x).

Let F(u)=2∫u t^2dt

g(x)=sinx then h(x)=F(g(x))

To differentiate use chain rule

From FTC2 F’(u)=u^2

g’(x)=cos(x)

h’(x)=F’(g(x))g’(x)= (sinx)^2cos(x)

In general,

(a∫g(x) f(t)dt)’ =f(g(x))g’(x)

(h(x)∫bf(t)dt)’ = -f(h(x))h’(x)

The average value of a function f(x) over the interval a≤x≤b by

1/(b-a) a∫bf(x)dx

Mean Value Theorem for Integrals

If f is continuous on a≤x≤b there is a number c between a and b such that

1/(b-a)a∫bf(x)dx=f(c)

i.e. f(c)=average value of f on interval a≤x≤b.

Area under a Curve

The area between the graph of y = f(x) and the x-axis is given by the definite integral below. This formula gives a positive result for a graph above the x-axis, and a negative result for a graph below the x-axis.

Note: If the graph of y = f(x) is partly above and partly below the x-axis, the formula given below generates the net area. That is, the area above the axis minus the area below the axis.

Formula: a∫b f(x)dx

Area between Curves

The area between curves is given by the formulas below:

Formula 1: Area=a∫b|f(x)-g(x)|dx

for a region bounded above and below by y = f(x) and y = g(x), and on the left and right by x = a and x = b.

Formula 2: Area=c∫d |f(y)-g(y)|dy

for a region bounded left and right by x = f(y) and x = g(y), and above and below by y = c and y = d.

Example1: Find the area between y = x and y = x2 from x = 1 to x = 2.

Example 2: Find the area between x = y + 3 and x = y2 from y = –1 to y = 1.

Method of slicing

A solid is sliced into n thin slices of equal width ∆x

Volume V = ∑V(xi) ≈ ∑A(xi)∆x

The actual volume should be the limit as ∆x→0

Volume V = limn→∞∑A(xi)∆x= a∫bA(x)dx

Example

Find volume of y=(r2-x2)1/2 rotated about the x-axis

A(x)=pi*y2

V=-r∫r A(x)dx = -r∫r pi*(r2-x2)dx

=pi(r2x-x3/3)-r|r

=4pi*r3/3

Cylindrical Shells Method

If the cross sections of the solid are taken parallel to the axis of revolution, then the cylindrical shell method will be used to find the volume of the solid. If the cylindrical shell has radius r and height h, then its volume would be 2π rh times its thickness. Think of the first part of this product, (2π rh), as the area of the rectangle formed by cutting the shell perpendicular to its radius and laying it out flat. If the axis of revolution is vertical, then the radius and height should be expressed in terms of x. If, however, the axis of revolution is horizontal, then the radius and height should be expressed in terms of y.

The volume ( V) of a solid generated by revolving the region bounded by y = f(x) and the x-axis on the interval [ a,b], where f(x) ≥ 0, about the y-axis is

If the region bounded by x = f(y) and the y-axis on the interval [ a,b], where f(y) ≥ 0, is revolved about the x-axis, then its volume ( V) is

Note that the x and y in the integrands represent the radii of the cylindrical shells or the distance between the cylindrical shell and the axis of revolution. The f(x) and f(y) factors represent the heights of the cylindrical shells.

Example: Find the volume of the solid generated by revolving the region bounded by y = x2 and the x-axis [1,3] about the y-axis.

In using the cylindrical shell method, the integral should be expressed in terms of x because the axis of revolution is vertical. The radius of the shell is x, and the height of the shell is f(x) = x2

The length of an arc along a portion of a curve is another application of the definite integral. The function and its derivative must both be continuous on the closed interval being considered for such an arc length to be guaranteed. If y = f(x) and y′ = F'(x) are continuous on the closed interval [ a, b], then the arc length ( L) of f(x) on [ a,b] is

Similarly, if x = f(y) and x' = f'( y) are continuous on the closed interval [ a,b], then the arc length ( L) of f(y) on [ a,b] is

Example: Find the arc length of the graph of on the interval [0,5].

Work=Force*Distance

For a constant force F moving an object from x=a to x=b,

W=F(b-a)

Suppose force varies continuously with position, then

W=a∫bF(x)dx

Example: A cable hangs vertically from the top of a building so that a length of 100m (which has a mass of 300kg) is hanging from the roof. Find the work required to lift the cable to the roof.

Let y=0 be the initial position of bottom of cable. Divide cable in to small pieces. Consider work done on typical piece with height dy located y above the bottom of the cable

F=Weight=(mass of piece)*g

=(length of piece)*(300kg/100m)*g= 3gdy

Work done to lift one piece= 3gdy*(100-y)

Work to lift cable= 0∫100 3gdy*(100-y) =3g(100y-y2/2)0|100

=147150 J

Centroidof a thin plate with uniform density ρ

The center of mass or centroid of a region is the point in which the region will be perfectly balanced horizontally if suspended from that point.

So, let’s suppose that the plate is the region bounded by the two curves f(x) and g(x) on the interval [a,b].

We’ll first need the mass of this plate. The mass is,

m= ρA= ρa∫bf(x)-g(x)dx

Next we’ll need the moments of the region. There are two moments, denoted by Mx and My. The moments measure the tendency of the region to rotate about the x and y-axis respectively. The moments are given by,

The coordinates of the center of mass are then,

Example: Determine the center of mass for the region bounded by y=2sin(2x) and x-axis on the interval [0, pi/2]

Let’s first get the area of the region,

Now, the moments (without density since it will just drop out) are,

The coordinates of the center of mass are then,

Good Luck On Your Final!

Zahra MahmoodBodla