Saturday

April 19, 2014

April 19, 2014

Number of results: 28,055

**calculus *improper integrals***

I see how it makes sense now, thanks a lot Steve!
*Sunday, September 16, 2012 at 11:57pm by John*

**Calculus**

What is the connection between improper integrals, Riemann sums, and the integral test?
*Tuesday, March 26, 2013 at 5:58pm by Kendra*

**calculus *improper integrals***

yes , the hole goes through the sphere, centered along the diameter
*Sunday, September 16, 2012 at 11:57pm by John*

**College Calculus**

Could someone explain how to distinguish improper integrals in non-mathematical terms so I may understand? Thanks!
*Thursday, February 28, 2008 at 7:59pm by Matt*

**calculus**

how would you do this improper integral 1/(x-1) from 0 to 2 this is improper at one, so I split it up into two integrals ln(x-1) from 0-1 and ln(x-1) from 1-2 I then did for the first one the (lim t->1(-) of ln(t-1))-(ln(0-1)) and then the same thing for the second part I ...
*Saturday, February 9, 2008 at 5:34pm by sarah*

**calculus**

how would you do this improper integral 1/(x-1) from 0 to 2 this is improper at one, so I split it up into two integrals ln(x-1) from 0-1 and ln(x-1) from 1-2 I then did for the first one the (lim t->1(-) of ln(t-1))-(ln(0-1)) and then the same thing for the second part I ...
*Sunday, February 10, 2008 at 10:54pm by sarah*

**Calculus ll - Improper Integrals**

Oh! I understand now. The "total area" part kind of threw me off. Thank you so much!
*Saturday, October 2, 2010 at 9:32pm by Alyssa*

**calculus *improper integrals***

The answer depends upon where the hole is drilled. Is it centered along a diameter of the sphere?
*Sunday, September 16, 2012 at 11:57pm by drwls*

**College Calculus**

Improper integrals implicitely imply taking limits of the upper and/or lower limits of the integration.
*Thursday, February 28, 2008 at 7:59pm by Count Iblis*

**Calculus ll - Improper Integrals**

or look at your expression of 1/2 - 1/(2t^2) as t ---> infinity, doesn't 1/(2t^2) approach 0 ? so you are left with 1/2 -0 = 1/2 , as your approximations suggested.
*Saturday, October 2, 2010 at 9:32pm by Reiny*

**Calculus (improper integrals)**

the integral from 0 to lnx of lnx/(x^1/2) thanks!
*Thursday, March 6, 2008 at 6:25pm by Chelsea*

**Calculus **

I do not know how you got 0.7, negative infinity seems to me correct because there is an asymptote at x=5. For improper integrals where the vertical asymptote is between the integration limites, we have to be careful not just evaluate the integral at the two limits. Instead, ...
*Friday, October 22, 2010 at 8:34pm by MathMate*

**calc**

how would you do this improper integral 1/(x-1) from 0 to 2 this is improper at one, so I split it up into two integrals ln(x-1) from 0-1 and ln(x-1) from 1-2 I then did for the first one the (lim t->1(-) of ln(t-1))-(ln(0-1)) and then the same thing for the second part I ...
*Saturday, February 9, 2008 at 6:34pm by sarah*

**Calculus ll - Improper Integrals**

area = LIM [integral]1/x^3 dx from x = 1 to x -- infin. = lim (-(1/2)(1/x^2) from x=1 to x-- inf. = lim{ -(1/2)/x^2 as x--inf} - lim {(-1/2)/x^2 as x=1 = 0 - (-1/2) = 1/2
*Saturday, October 2, 2010 at 9:32pm by Reiny*

**calculus**

[integrals]2/tsqrt(t^4+25) integrals of two over t square root of t to the 4th plus 25
*Monday, November 23, 2009 at 1:26am by steve*

**Calculus (Definite Integrals)**

How many definite integrals would be required to represent the area of the region enclosed by the curves y=(cos^2(x))(sin(x)) and y=0.03x^2, assuming you could not use the absolute value function? a.) 1 b.) 2 c.) 3 d.) 4 e.) 5
*Wednesday, February 29, 2012 at 7:55pm by Mishaka*

**calculus**

Unless you meant 1 + x^4 in the denominator or x^4 in the numerator, that integral is a very difficult one to do. Are you sure you typed g(x) correctly? There is a recursion formula for that integral in my Table of Integrals, but it leads to two new integrals that look even ...
*Sunday, February 17, 2008 at 11:25pm by drwls*

**Calculus AB**

Consider the region bounded by the graphs of the equations x=y^2 and y=3x. Set up 2 integrals, one with respect to x and the other with respect to y, both of which compute the volume of the solid obtained by rotating this region about the x-axis and evaluate the integrals.
*Tuesday, March 8, 2011 at 1:10pm by Nellie*

**calculus *improper integrals***

A hole of a radius of 1cm is pierced in a sphere of a 4cm radius. Calculate the volume of the remaining sphere. I know that the volume of a sphere with an exterior radius is of: V = pie(R^2 - r^2) dH where R^2 is the exterior surface area and r^2 is the interior surface area. ...
*Sunday, September 16, 2012 at 11:57pm by John*

**calculus**

There are four integrals: 1) definite integral x/(1+x^4)dx b/w 0_infinity 2) definite integral (x^2)/(1+x^4)dx b/w 0_infinity 3) definite integral (x^3)/(1+x^4)dx b/w 0_infinity 4) definite integral (x^4)/(1+x^4)dx b/w 0_infinity Which of these integrals converge. First of all...
*Wednesday, March 3, 2010 at 4:24pm by Carmen*

**calculus**

There are four integrals: 1) definite integral x/(1+x^4)dx b/w 0_infinity 2) definite integral (x^2)/(1+x^4)dx b/w 0_infinity 3) definite integral (x^3)/(1+x^4)dx b/w 0_infinity 4) definite integral (x^4)/(1+x^4)dx b/w 0_infinity Which of these integrals converge. First of all...
*Thursday, March 4, 2010 at 2:03am by Carmen*

**calculus *improper integrals***

So far so good. h^2 = R^2 - x^2 2h dh = -2x dx or, dh = -x/h dx so the integral becomes pi (R^2-r^2) (-x/sqrt(R^2 - x^2)) dx = -pi (R^2-r^2) x/sqrt(R^2-x^2) dx now let u = R^2 - x^2 du = -2x dx and the integral is pi/2 (R^2-r^2) u^(-1/2) du If I've done things right, v = 4/3 ...
*Sunday, September 16, 2012 at 11:57pm by Steve*

**MATH 2B Calculus **

Consider the area between the graphs x+4y=14 and x+7=y^2. This area can be computed in two different ways using integrals. First of all it can be computed as a sum of two integrals They ask to use two integrals so i put f(x) from -7 to 2 which is correct but for g(x) i put 2 ...
*Friday, August 20, 2010 at 10:58pm by TOMO*

**Calculus - evaluating integrals**

I'm really having trouble with this current topic that we're learning. Any explanations are greatly appreciated. Evaluate the integrals: 1.) ∫ (2-2cos^2x) dx 2.) ∫ cot3x dx 3.) ∫ ((e^(sqrt x) / (sqrt x) dx))
*Wednesday, March 23, 2011 at 11:40am by Amy*

**Calculus - Integrals (part 1)**

<<Integral of x sqrt(19x-7)dx >> Let u = 19x -7 dx = (1/19) du x = (1/19)(u + 7) So the integral becomes the sum of two integrals: Integral of (1/19)u^(3/2) du + Integral of (7/19)u^(1/2) du both of which are easy integrals. Remember to change from u back to (19x...
*Sunday, March 16, 2008 at 11:22am by drwls*

**Calculus**

No, not true for all integrals. But if you are looking for something that only has magnitude, you have to split the integrals, as the area below the axis is NEGATIVE. On things like vector work (force*dx), the negative would mean work being absorbed, so it might be useful to ...
*Friday, December 18, 2009 at 1:56pm by bobpursley*

**Calculus Area between curves**

Consider the area between the graphs x+6y=8 and x+8=y2. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals where a= , b=, c= and f(x)= g(x)= I found a, but not b or c. I can't seem to figure out f(x) and g(...
*Sunday, November 24, 2013 at 10:20pm by Kelly*

**calculus**

You used a correct procedure, but both integrals you broke it up into involve the log of zero or a negative number, both of which are "improper". I don't see how one can get a finite integral this way. Here is another way that might work: Define the integral of the sum of two ...
*Saturday, February 9, 2008 at 5:34pm by drwls*

**Calculus**

9(sinx + cosx)/sin2x = 9( sinx/(2sinxcosx) + cosx/(2sinxcosx) = (9/2)(1/cosx + 1/sinx) = (9/2) (secx + cscs) can you take it from there? (the integrals of secx and cscx should be part of your repertoire of basic integrals)
*Thursday, March 7, 2013 at 12:22am by Reiny*

**MATH**

Evaluate by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area.
*Friday, December 7, 2012 at 5:03am by LUNG*

**MATH**

mixed numbers are pretty well useless in actual calculations, since the first thing one usually has to do is change them to improper fractions. I also resent the fact that something like 5/2 is called an "improper" fraction, there is nothing improper about them. I never really...
*Wednesday, July 8, 2009 at 11:39am by Reiny*

**Calculus (Integrals)**

56
*Wednesday, February 15, 2012 at 7:41pm by anne*

**Calculus (Integrals)**

56
*Wednesday, February 15, 2012 at 7:41pm by anne*

**Calculus**

Integrals: When we solve for area under a curve, we must consider when the curve is under the axis. We would have to split the integral using the zeros that intersect with the axis. Would this be for all integrals? What if we just want to "find the integral", without finding ...
*Friday, December 18, 2009 at 1:56pm by Jennifer*

**Calculus ll - Improper Integrals**

Find the area of the curve y = 1/(x^3) from x = 1 to x = t and evaluate it for t = 10, 100, and 1000. Then find the the total area under this curve for x ≥ 1. I'm not sure how to do the last part of question ("find the the total area under this curve for x ≥ 1.") ...
*Saturday, October 2, 2010 at 9:32pm by Alyssa*

**improper integral**

The indefinite integral is -2 e^(-sqrt(t)) At t = 0, the value of the indefinite integral is -2, and at t = infinity, it approaches zero Therefore the integral is 0 - (-2) = 2 Integrals to not have to diverge just because the integrand diverges at the endpoints. It all depends...
*Sunday, May 2, 2010 at 2:07am by drwls*

**Calculus - Integrals**

What is the integral of arctan x?
*Thursday, March 13, 2008 at 9:43pm by Sean*

**Calculus - Integrals**

1) A/x + B/(1+x) 2) A/x + B/x^2 + C/(1+x) I think I'm right....
*Monday, March 24, 2008 at 12:53pm by David*

**Calculus (integrals)**

20.25
*Friday, April 16, 2010 at 6:52pm by Naumair*

**Mixed numbers and improper fractions**

What's the improper fraction of 4 24/5
*Wednesday, January 25, 2012 at 4:38pm by Sam*

**Calculus**

integral -oo, oo [(2x)/(x^2+1)^2] dx (a) state why the integral is improper or involves improper integral *infinite limit of integration (b) determine whether the integral converges or diverges converges? (c) evaluate the integral if it converges I know f(x)=arctan->f'(x)=1...
*Friday, February 8, 2008 at 9:48pm by Anonymous*

**Calculus - Integrals**

O_O OH DEAR GOD.
*Tuesday, March 25, 2008 at 9:43pm by mclovin'*

**Calculus**

[Integrals] h(x)= -4 to sin(x) (cos(t^5)+t)dt h'(x)=?
*Saturday, November 21, 2009 at 7:48pm by Z32*

**Calculus (Integrals)**

Actually none of the above
*Wednesday, February 15, 2012 at 7:41pm by anne*

**Calculus - Integrals**

Thank you so much. You cleared that up really well. Thank you.
*Sunday, March 16, 2008 at 11:22am by Sean*

**Calculus - Integrals**

You now just need to solve for A, B, and C :)
*Monday, March 24, 2008 at 12:53pm by Count Iblis*

**calculus 4**

Why use triple integrals? The CM is at the average x, y and z: x = 0, z=0 and y=4.
*Tuesday, November 3, 2009 at 3:00pm by drwls*

**calculus**

is it because every interval of one the integrals approach 1?
*Wednesday, February 24, 2010 at 11:13am by Paul*

**calculus 2**

integrate x^5/(x^2 + sqrt(2)) using a table of integrals
*Wednesday, March 30, 2011 at 12:27am by Dana*

**Calculus**

It's the same as the integral of x^-2 + x^-5 Add the integrals if each term.
*Monday, October 31, 2011 at 7:46pm by drwls*

**Calculus II**

Could you show how to do this problem using integrals?
*Monday, January 23, 2012 at 8:14pm by Morgan*

**Calculus - Integrals (part 2)**

<<Integral of [(sin(7x)^2)*(sec(7x)^4) dt] >> I assume your differential variable of integration is dx, not dt. I suggest you rewrite sin(7x)^2 as 1 = cos^2(7x). That leaves you with two integrals: one involving sec^4(7x) and the other sec^2(7x) (since cos = 1/sec...
*Sunday, March 16, 2008 at 11:22am by drwls*

**Calculus - Integrals**

I don't quite get how you did the first one. Mainly the cos(1/2 x)
*Friday, March 14, 2008 at 9:00pm by Sean*

**single variable calculus - indefinite integrals**

Oh! I didn't realize it was so simple. Thank you!
*Thursday, August 11, 2011 at 7:53pm by Kelly*

**math**

1. change to improper fraction: 9 5/9 2. change to improper fraction: 6 23/39 3. change the improper fraction to a whole number:115/20 4. multiply 4 3/4 x 1 1/7 5,subtract- 20 11/15- 17 1/3 6 2(-6)(-8)(0)(-5): multiply 7. solve for the variable: 0.1p=7
*Tuesday, May 14, 2013 at 12:45am by John*

**improper fraction**

improper fractions like 4 2/3
*Thursday, January 20, 2011 at 8:31am by Anonymous*

**Maths**

What is the answer for these questions:- 1) Indefinite Integrals gcx) = (8 + 39x ^ 3) / x 2) Indefinite Integrals hcu) = sin ^2 (1/8 u) 3) Evaluate x ( 8 - 5 x ^2) dx Thank you
*Thursday, June 16, 2011 at 4:49am by Alex*

**Calc or Pre calc**

I am having trouble doing this problem. I know how to do indefinate integrals, but I don't know how to do definate integrals. Can you show me how to do this. Evaluate 5 (x^3-2x)dx 2
*Wednesday, May 14, 2008 at 9:42pm by jennifer*

**Calculus**

straightforward power integrals. What do you get? ∫12x^2 dx = 4x^3, etc.
*Wednesday, November 21, 2012 at 12:05am by Steve*

**Calculus**

Hm, I've never done integrals using that method before. I'll check it.
*Wednesday, January 9, 2013 at 7:48pm by Kyle*

**calculus **

1. break it into three integrals INT (x dx)+int(7x^-2 dx) - int x^-8dx 2.multipy it out break it into three inegrals, and use the power rule. 3. two integrals, power rule.
*Sunday, July 25, 2010 at 7:45pm by bobpursley*

**calculus**

Use integrals to prove that the volume of a sphere of radius R is equal to (4/3)(pi)R^3
*Saturday, January 31, 2009 at 4:53pm by matt*

**Calculus**

definite Integrals (using fundamental Theorem) Evaluate from -1 to 2(x^2 - 4x)dx
*Friday, December 2, 2011 at 1:10am by Stacy *

**Calculus - Integrals**

Sorry. I lied. Misread the question. What is the indefinite integral of arccot(4x) ?
*Thursday, March 13, 2008 at 9:43pm by Sean*

**Calculus - Integrals**

Integral of (dt/[sqrt(t)+25tsqrt(t)]} From 1/75 to 3/25 I'm at a complete loss at what to do.
*Saturday, March 15, 2008 at 9:55pm by Sean*

**Calculus - Integrals**

Could you expand on #1 a bit more? I tried Partial Fractions, but I couldn't get a definite answer...
*Sunday, March 23, 2008 at 12:12pm by David*

**Calculus - Partial Integrals**

I'm confused. As soon as I get the so called 'Basic Equation,' I just don't know what to do after....
*Tuesday, March 25, 2008 at 7:19am by David*

**Calculus**

Graph the integrands and use areas to evaluate integrals (integrate(3&-3)) root(9-x^2) dx
*Friday, January 24, 2014 at 10:34pm by Anonymous*

**math**

1) convert the mixed number into an improper fraction http://www.ehow.com/how_4546598_convert-mixed-numbers-improper-fractions.html 2) multiply 4/5 to the improper fraction
*Wednesday, October 20, 2010 at 1:21am by TutorCat*

**Calculus**

Hi. How can I integrate 1/(X^3 +1) ? Thank you to anyone who can help me :-) Write 1/(x^3 +1) as 1/[(x+1)(x^2-x+1)] Then use integration by parts, letting dv = dx/(x^2 -x +1) u = 1/(x+1) du = log (x+1) v = (2/sqrt3)arctan[(2x-1)/sqrt3] That should take you to the answer. I ...
*Sunday, March 18, 2007 at 10:52pm by M*

**math**

Since it didnt ask for an improper fraction or mixed number and that is what we are studying I came up with 40/8 as the improper and 5 for the mixed?
*Wednesday, February 8, 2012 at 11:53am by Seth*

**Mixed numbers and improper fractions**

Improper fraction as a mixed number in simplest form? 29/18
*Wednesday, January 25, 2012 at 4:38pm by Shamekka teart*

**math, calculus 2 **

Consider the area between the graphs x+y=16 and x+4= (y^2). This area can be computed in two different ways using integrals. First of all it can be computed as a sum of two integrals integrate from a to b of f(x)dx + integrate from b to c of g(x)dx What is the value of a, b, c...
*Wednesday, September 12, 2012 at 8:03am by bobby*

**math**

The integral of y/cos^2y dy = The integral of e*e^x dx + Constant y tany + log(cosy) = e^x*(x-1) + C Use an initial condition to evaluate C Verify the integrals yourself. I used a table of integrals.
*Wednesday, March 30, 2011 at 3:54am by drwls*

**calculus**

evaluate the improper integral whenever it is convergent ∫0,-00 1/(4-x)^(3/2) dx
*Sunday, March 4, 2012 at 11:43am by help please.*

**Calculus - Partial Integrals**

The paragraph after the thing with s as a variable is copy and pasted from Count Iblis.
*Tuesday, March 25, 2008 at 7:19am by David*

**Calculus - Partial Integrals**

The paragraph after the thing with s as a variable is copy and pasted from Count Iblis.
*Tuesday, March 25, 2008 at 7:19am by David*

**calculus**

Use the Table of Integrals to evaluate the integral (x sine(6x^2)cos(7x^2)dx)
*Sunday, February 14, 2010 at 11:28pm by ali*

**single variable calculus - indefinite integrals**

integral of (1-(sinx)^2))/(cosx)dx i don't know what to make my "u" for u-substitution
*Thursday, August 11, 2011 at 7:53pm by Kelly*

**Constitutional Law**

is the exclusion of evidence that is obtained through an improper search the appropriate response to an improper search or is there a better way to deal with such situation
*Wednesday, April 18, 2012 at 11:51am by Stacy*

**math**

12 1/3 + 7 7/15 switch fractions to improper fractions then find a common denominator to find improper fraction of 12 1/3 multiply whole number (12) by the denominator of the fraction (3) then add the numerator of the fraction (1) therefore improper fraction will be 37/3
*Wednesday, April 25, 2007 at 6:32pm by Freddie*

**calculus (please help steve)**

Indeed. I have misread the question and just calculated the area. Some work will need to be done on the function F(x,y)=(x^2+y^2)^2/(x^2*y^2) because it is an improper integral as (x,y)->(0,0). The existence of the integral will have to be demonstrated at (0,0), and the ...
*Thursday, July 25, 2013 at 6:17pm by MathMate*

**Calculus - Integrals**

I'm not quite sure about your explanation on Partial Fractions... And Liouville's Therom. We haven't done that in class... so...
*Monday, March 24, 2008 at 12:53pm by David*

**calculus**

evaluate the following indefinite integrals by substitution & check the result by differentiation. ∫(sin2x)^2 cos2xdx
*Thursday, December 8, 2011 at 12:29am by Andresito*

**calculus**

Evaluate the following integrals by using appropriate method : ∫cos ^3 ( 2x-5 )dx help
*Sunday, December 25, 2011 at 12:31pm by Jacob*

**calculus**

much too nasty to work out here, try http://integrals.wolfram.com/index.jsp?expr=sqrt(1%2Bx^3)&random=false
*Monday, February 23, 2009 at 11:22am by Reiny*

**cal**

One of the integrals you should have in your repertoire of common integrals is ∫lnx = xlnx - x so volume = π∫ lnx dx from x-1 to e^2 = xln - x | from 1 to e^2 = e^2(lne^2) - e^2 - (1ln1 - 1) = e^2(2) - e^2 - 0 + 1 = e^2 + 1
*Monday, December 23, 2013 at 10:24am by Reiny*

**calculus 4**

a solid cube, 2 units on a side, is bounded by the planes x=+-1, z=+-1, y=3 and y=5. Find the center of mass using triple integrals.
*Tuesday, November 3, 2009 at 3:00pm by john*

**Calculus**

For these kind of integrals you either have to go to tables of integrals or find a suitable computer program I found the integral of √(1+t^2) to be [ln(√(1+t^2) + t)]/2 + [t√(1+t^2)]/2 after evaluating this from 1 to x^3 I got [ln(√(1+x^6) + x^3)]/2 + [...
*Wednesday, December 5, 2007 at 3:16am by Reiny*

**AP Calculus**

how do you evaluate integral of [lxl] from (0,2) , a step function. would you just do it as if it were a absolute value and then do two different integrals?
*Tuesday, February 10, 2009 at 8:24pm by David*

**calculus**

Is the second term (4/7)x or 4/(7x) ? It makes a big differnece. Add the separate integrals of the two terms. The indefinite integral of 4/x^5 (4 x^-5) is 4 x^-4/(-4) = -x^(-4)
*Sunday, December 13, 2009 at 12:37am by drwls*

**Calculus**

Integrate [1/square root of(e^(2x)-1)]. I have to use u substitution. We are doing the integrals of inverse trig functions, but I cannot get it to work out!
*Sunday, February 5, 2012 at 10:43pm by Elaine*

**calculus**

i just didn't understand how you got to that point. And i havn't learned integrals yet. I was hoping someone else could give me a different perspective as to how to answer it.
*Sunday, November 8, 2009 at 6:10pm by Reen*

**Calculus - Urgent!!!**

Ah, thank you so much. Those are very helpful. Yes, anti-derivatives and integrals are in my text, but it is explained in one paragraph. Thanks again!
*Thursday, April 26, 2012 at 5:18pm by In Need Of Help!*

**calculus**

I "cheated" I went to http://integrals.wolfram.com/index.jsp?expr=ln(x^2+-+x+%2B+2)&random=false
*Thursday, February 5, 2009 at 11:53pm by Reiny*

**structurlal design 2**

15. Sagging in floor joists is typically caused by A. oversized joists. B. undersized joists. C. improper nailing of sub-flooring. D. improper moisture content of the wood i think it's D.
*Sunday, February 24, 2013 at 5:03am by Austin*

**calculus II university**

prove the following integrals: a)sin3xcos7xdx = -1/20cos(10x)+1/8cos(4x) b)sin8xcos3xdx = -1/22cos(11x)-1/10cos(5x)
*Thursday, January 28, 2010 at 7:24am by maria*

**Calculus**

"Evaluate the following indefinite integrals: "S" (3x^2 -2)/(x^3 - 2x + 1)^3 dx" We're practicing the substitution rule, and I know how to do it, but I don't know what/how to substitute in this question. btw: "S" is the integral sign.
*Sunday, April 11, 2010 at 11:19pm by Stuck*

**calculus**

set up sums of integrals that can be used to find the area of the region bounded by the graphs of the equations by integrating with respect to y y=square root of x; y= -x, x=1, x=4
*Tuesday, May 1, 2012 at 10:27am by dario*

**Calculus (integrals)**

it surely is wrong If u = x^8 + 6x, then you have du/u^2 integral is -1/u try taking the derivative of ln(u^2) and see what you get. Don't forget the chain rule.
*Tuesday, October 29, 2013 at 10:21pm by Steve*

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