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f ( x ) = sin x , a = 0 , b = 2 π

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f ( x ) = cos x , a = 0 , b = 2 π

f ave = 0 ; c = π 2 , 3 π 2

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In the following exercises, approximate the average value using Riemann sums L 100 and R 100 . How does your answer compare with the exact given answer?

[T] y = ln ( x ) over the interval [ 1 , 4 ] ; the exact solution is ln ( 256 ) 3 1 .

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[T] y = e x / 2 over the interval [ 0 , 1 ] ; the exact solution is 2 ( e 1 ) .

L 100 = 1.294 , R 100 = 1.301 ; the exact average is between these values.

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[T] y = tan x over the interval [ 0 , π 4 ] ; the exact solution is 2 ln ( 2 ) π .

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[T] y = x + 1 4 x 2 over the interval [ −1 , 1 ] ; the exact solution is π 6 .

L 100 × ( 1 2 ) = 0.5178 , R 100 × ( 1 2 ) = 0.5294

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In the following exercises, compute the average value using the left Riemann sums L N for N = 1 , 10 , 100 . How does the accuracy compare with the given exact value?

[T] y = x 2 4 over the interval [ 0 , 2 ] ; the exact solution is 8 3 .

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[T] y = x e x 2 over the interval [ 0 , 2 ] ; the exact solution is 1 4 ( e 4 1 ) .

L 1 = 0 , L 10 × ( 1 2 ) = 8.743493 , L 100 × ( 1 2 ) = 12.861728 . The exact answer 26.799 , so L 100 is not accurate.

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[T] y = ( 1 2 ) x over the interval [ 0 , 4 ] ; the exact solution is 15 64 ln ( 2 ) .

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[T] y = x sin ( x 2 ) over the interval [ π , 0 ] ; the exact solution is cos ( π 2 ) 1 2 π .

L 1 × ( 1 π ) = 1.352 , L 10 × ( 1 π ) = −0.1837 , L 100 × ( 1 π ) = −0.2956 . The exact answer 0.303 , so L 100 is not accurate to first decimal.

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Suppose that A = 0 2 π sin 2 t d t and B = 0 2 π cos 2 t d t . Show that A + B = 2 π and A = B .

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Suppose that A = π / 4 π / 4 sec 2 t d t = π and B = π / 4 π / 4 tan 2 t d t . Show that A B = π 2 .

Use tan 2 θ + 1 = sec 2 θ . Then, B A = π / 4 π / 4 1 d x = π 2 .

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Show that the average value of sin 2 t over [ 0 , 2 π ] is equal to 1/2 Without further calculation, determine whether the average value of sin 2 t over [ 0 , π ] is also equal to 1/2.

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Show that the average value of cos 2 t over [ 0 , 2 π ] is equal to 1 / 2 . Without further calculation, determine whether the average value of cos 2 ( t ) over [ 0 , π ] is also equal to 1 / 2 .

0 2 π cos 2 t d t = π , so divide by the length 2 π of the interval. cos 2 t has period π , so yes, it is true.

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Explain why the graphs of a quadratic function (parabola) p ( x ) and a linear function ( x ) can intersect in at most two points. Suppose that p ( a ) = ( a ) and p ( b ) = ( b ) , and that a b p ( t ) d t > a b ( t ) d t . Explain why c d p ( t ) > c d ( t ) d t whenever a c < d b .

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Suppose that parabola p ( x ) = a x 2 + b x + c opens downward ( a < 0 ) and has a vertex of y = b 2 a > 0 . For which interval [ A , B ] is A B ( a x 2 + b x + c ) d x as large as possible?

The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, A = b b 2 4 a c 2 a and B = b + b 2 4 a c 2 a .

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Suppose [ a , b ] can be subdivided into subintervals a = a 0 < a 1 < a 2 < < a N = b such that either f 0 over [ a i 1 , a i ] or f 0 over [ a i 1 , a i ] . Set A i = a i 1 a i f ( t ) d t .

  1. Explain why a b f ( t ) d t = A 1 + A 2 + + A N .
  2. Then, explain why | a b f ( t ) d t | a b | f ( t ) | d t .
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Suppose f and g are continuous functions such that c d f ( t ) d t c d g ( t ) d t for every subinterval [ c , d ] of [ a , b ] . Explain why f ( x ) g ( x ) for all values of x .

If f ( t 0 ) > g ( t 0 ) for some t 0 [ a , b ] , then since f g is continuous, there is an interval containing t 0 such that f ( t ) > g ( t ) over the interval [ c , d ] , and then d d f ( t ) d t > c d g ( t ) d t over this interval.

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Suppose the average value of f over [ a , b ] is 1 and the average value of f over [ b , c ] is 1 where a < c < b . Show that the average value of f over [ a , c ] is also 1.

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Suppose that [ a , b ] can be partitioned. taking a = a 0 < a 1 < < a N = b such that the average value of f over each subinterval [ a i 1 , a i ] = 1 is equal to 1 for each i = 1 ,…, N . Explain why the average value of f over [ a , b ] is also equal to 1.

The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, a b f ( t ) d t = a 0 a 1 f ( t ) d t + a 1 a 2 f ( t ) d t + + a N + 1 a N f ( t ) d t = a 0 a 1 1 d t + a 1 a 2 1 d t + + a N + 1 a N 1 d t = ( a 1 a 0 ) + ( a 2 a 1 ) + + ( a N a N 1 ) = a N a 0 = b a . Dividing through by b a gives the desired identity.

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Suppose that for each i such that 1 i N one has i 1 i f ( t ) d t = i . Show that 0 N f ( t ) d t = N ( N + 1 ) 2 .

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Suppose that for each i such that 1 i N one has i 1 i f ( t ) d t = i 2 . Show that 0 N f ( t ) d t = N ( N + 1 ) ( 2 N + 1 ) 6 .

0 N f ( t ) d t = i = 1 N i 1 i f ( t ) d t = i = 1 N i 2 = N ( N + 1 ) ( 2 N + 1 ) 6

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[T] Compute the left and right Riemann sums L 10 and R 10 and their average L 10 + R 10 2 for f ( t ) = t 2 over [ 0 , 1 ] . Given that 0 1 t 2 d t = 0. 33 , to how many decimal places is L 10 + R 10 2 accurate?

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[T] Compute the left and right Riemann sums, L 10 and R 10 , and their average L 10 + R 10 2 for f ( t ) = ( 4 t 2 ) over [ 1 , 2 ] . Given that 1 2 ( 4 t 2 ) d t = 1. 66 , to how many decimal places is L 10 + R 10 2 accurate?

L 10 = 1.815 , R 10 = 1.515 , L 10 + R 10 2 = 1.665 , so the estimate is accurate to two decimal places.

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If 1 5 1 + t 4 d t = 41.7133.. . , what is 1 5 1 + u 4 d u ?

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Estimate 0 1 t d t using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value 0 1 t d t ?

The average is 1 / 2 , which is equal to the integral in this case.

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Estimate 0 1 t d t by comparison with the area of a single rectangle with height equal to the value of t at the midpoint t = 1 2 . How does this midpoint estimate compare with the actual value 0 1 t d t ?

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From the graph of sin ( 2 π x ) shown:

  1. Explain why 0 1 sin ( 2 π t ) d t = 0 .
  2. Explain why, in general, a a + 1 sin ( 2 π t ) d t = 0 for any value of a .
    A graph of the function f(x) = sin(2pi*x) over [0, 2]. The function is shaded over [.7, 1] above the curve and below to x axis, over [1,1.5] under the curve and above the x axis, and over [1.5, 1.7] above the curve and under the x axis. The graph is antisymmetric with respect o t = ½ over [0,1].

a. The graph is antisymmetric with respect to t = 1 2 over [ 0 , 1 ] , so the average value is zero. b. For any value of a , the graph between [ a , a + 1 ] is a shift of the graph over [ 0 , 1 ] , so the net areas above and below the axis do not change and the average remains zero.

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If f is 1-periodic ( f ( t + 1 ) = f ( t ) ) , odd, and integrable over [ 0 , 1 ] , is it always true that 0 1 f ( t ) d t = 0 ?

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If f is 1-periodic and 0 1 f ( t ) d t = A , is it necessarily true that a 1 + a f ( t ) d t = A for all A ?

Yes, the integral over any interval of length 1 is the same.

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Questions & Answers

questions solve y=sin x
Obi Reply
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
Ryan Reply
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
NIKI Reply
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
K.kupar Reply
ask a complete question if you want a complete answer.
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
Darnell Reply
proof the formula integration of udv=uv-integration of vdu.?
Bg Reply
Find derivative (2x^3+6xy-4y^2)^2
Rasheed Reply
no x=2 is not a function, as there is nothing that's changing.
Vivek Reply
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
Joys Reply
y=800
Gift
800
Bg
how do u factor the numerator?
Drew Reply
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
follow algebraic method. look under factoring numerator from Khan academy
moe
volume between cone z=√(x^2+y^2) and plane z=2
Kranthi Reply
answer please?
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Lerato Reply
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
Amna Reply
what is a function? f(x)
Jeremy Reply
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
jon Reply
is x=2 a function?
The
Practice Key Terms 8

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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