Friday

April 18, 2014

April 18, 2014

Number of results: 27,478

**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**

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*

**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 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**

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**

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*

**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 (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**

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*

**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 (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 **

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*

**Calculus - Integrals**

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

**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*

**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*

**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*

**Calculus - Integrals**

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

**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*

**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 **

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**

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**

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 *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**

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, 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*

**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 *improper integrals***

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

**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**

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*

**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 - 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*

**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*

**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 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**

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 *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*

**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**

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*

**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**

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*

**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 use this page instead of looking integrals up on tables http://integrals.wolfram.com/index.jsp?expr=x%2F%28x%5E2%28x%5E4-1%29%5E%281%2F2%29%29&random=false here is the second one http://integrals.wolfram.com/index.jsp?expr=1%2F+%281%2B%28x%5E2%2F4%29%29&random=false
*Sunday, November 14, 2010 at 5:17pm by Reiny*

**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 need help with integrals and I need help with the problem. Integral of 4/sqrt(x)dx
*Monday, April 11, 2011 at 9:30pm by Leanna*

**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 showed work**

find the area of the rgion bounded by the graphs of y=x^3-2x and g(x)=-x i drew the graph and half of the graph is above the xaxis and the other half is below the axis. so the integrals i came up with are two because i broke them up and i combined the answers at the end: ...
*Saturday, January 6, 2007 at 9:05pm by david*

**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*

**Calculus - Integrals**

That's my problem. When I tried solving for A, B, C, D, and E in that first problem, it didn't really work out right...
*Monday, March 24, 2008 at 12:53pm by David*

**calculus**

evaluate the following integrals by any means possible: b. integral from -1 to 1 (w/ (w^2+1)) (dw) c. integral from -3 to 0 sqrt(9-x^2)dx
*Friday, July 3, 2009 at 9:16pm by Dora*

**Calculus (integrals)**

Find the integral:8x^7+6/(x^8+6x)^2 I got ln(x^8+6x)^2 but apparently that is wrong.
*Tuesday, October 29, 2013 at 10:21pm by Eve *

**Calculus (Integrals and Derivatives)**

It would be -28. For future notice, the integral is going from the interval -3 to 1(from point a to b)...this information will come in handy if you have to take physics
*Wednesday, February 15, 2012 at 6:58pm by john*

**Calculus **

Consider the area between the graphs x+1y=12 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 Interval a to b f(x)= i put 12-x but it's wrong Interval c to b f(x)= i put (x-8)^(1/2) but it's ...
*Sunday, April 29, 2012 at 10:32pm by jasmineT*

**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 **

1. Evaluate the following integrals (a) 4x2 +6x−12 / x3 − 4x dx
*Sunday, May 26, 2013 at 3:33am by Maryam*

**Calculus**

How on earth did you get that mess? Just add the separate integrals of 8x and -x^2. The answer is 4x^2 - (x^3)/3 To verify, differentiate it and see if you get 8x-x^2 back again.
*Friday, February 13, 2009 at 2:05am by drwls*

**single variable calculus - indefinite integrals**

ah (1-(sinx)^2)) = cos^2 x Unless I am mistaken or there is a typo all you have is cos x dx which is sin x + c
*Thursday, August 11, 2011 at 7:53pm by Damon*

**Cal**

consider the area between the graphs x+3y=1 and x+9=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 where a=,b=,c= and f(x0= and g(x)=. Alterntaively this area can be computed as a single integral
*Thursday, July 7, 2011 at 7:52pm by Lan*

**Cal**

consider the area between the graphs x+3y=1 and x+9=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 where a=,b=,c= and f(x0= and g(x)=. Alterntaively this area can be computed as a single integral
*Thursday, July 7, 2011 at 8:51pm by Lan*

**calculus 2**

Wolfram agrees with you, they took out a common factor of 63 http://integrals.wolfram.com/index.jsp?expr=63x%28cos%28x%29%29%5E2&random=false
*Friday, September 6, 2013 at 9:30pm by Reiny*

**Calculus**

Evaluate the following integrals {e˄x+e˄-x/ e˄x dx
*Monday, April 4, 2011 at 11:38am by Zukii*

**calculus**

prove the following integrals: a)sin3xcos7xdx = -1/20cos(10x)+1/8cos(4x) i know i have asked this question before but i am very confused bobpursley, can you please show me the steps of proves. thanks in advanced
*Thursday, January 28, 2010 at 2:24pm by maria*

**Calculus**

a) correct b) on your graph, x = -4 and x = 1 are vertical asymptotes. c) causes all kinds of headaches in such calculations as finding definite integrals, or even just making a sketch of the graph
*Saturday, February 11, 2012 at 1:46pm by Reiny*

**Calculus - Integrals**

I have 3 questions, and I cannot find method that actually solves them. 1) Integral [(4s+4)/([s^2+1]*([S-1]^3))] 2) Integral [ 2*sqrt[(1+cosx)/2]] 3) Integral [ 20*(sec(x))^4 Thanks in advance.
*Sunday, March 23, 2008 at 12:12pm by David*

**Calculus - Integrals**

I have 3 questions, and I cannot find method that actually solves them. 1) Integral [(4s+4)/([s^2+1]*([S-1]^3))] 2) Integral [ 2*sqrt[(1+cosx)/2]] 3) Integral [ 20*(sec(x))^4 Thanks in advance.
*Monday, March 24, 2008 at 12:53pm by David*

**calculus**

Do you mean lny/(6y^3) ? If so, the indefinite integral is [(-lny)/2 -(1/4)]/6y^2 I had to use a Table of Integrals to get that.
*Saturday, October 9, 2010 at 7:04pm by drwls*

**Calculus (Integrals)**

∫(2t^3 + 3) dt for t=-3 to 1 = [ (1/2)t^4 + 3t] from -3 to 1 = (1/2)(1) + 3 - ( (1/2)(81) -9) = 1/2 - 81/2 + 12 = -40 + 12 = -28
*Wednesday, February 15, 2012 at 7:41pm by Reiny*

**Calculus**

1 0 x^2dx=1/3. Use this and the properties of integrals to evaluate 1 0 10−2x^2dx
*Sunday, May 12, 2013 at 3:23am by Brian*

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