Sunday

April 20, 2014

April 20, 2014

Number of results: 5,198

**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*

**calc**

find integral using table of integrals ) integral sin^4xdx this the formula i used integral sin^n xdx =-1/n sin^n-1xcosx +n-1/n integral sin^n-2 using the formula this is what i got: integral sin^4xdx=-1/4sin^3xcosx+3/4 integral sin^2xdx= -1/2sinxcosx+1/2 integral 1 dx can ...
*Sunday, February 20, 2011 at 7:56pm by tom*

**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*

**calc**

It would be great to see how this is done. The prof's way of doing it is way confusing and intricate and I just started learning integrals.
*Tuesday, September 8, 2009 at 2:42am by Joe*

**AP Calc AB**

how do you use rules for integrals? in general, but I'm just looking for examples if possible...this is pretty new so I don't have a very good understanding yet. thanks.
*Monday, January 10, 2011 at 11:06pm by benches*

**calc ii**

I got it. its that the minimum area is found by saying that the function is a straight line at y=12 and the opposite for 18 and finding those areas under the integrals bounds
*Friday, September 27, 2013 at 8:13pm 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*

**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*

**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*

**Calc**

I did: sec^3x/tanx = (1/cos^3 x)(cosx/sinx) = 1/((sinx)(cos^2 x) so I went to my favourite integrator page and got http://integrals.wolfram.com/index.jsp?expr=1%2F%28%28sin%28x%29cos%5E2%28x%29%29&random=false You should realize that when Wolfram says log(...) they really mean...
*Monday, October 1, 2012 at 10:16pm by Reiny*

**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*

**Calc 2**

volume = π∫1/(x^2 + 2x + 5)^2 dx from 0 to 1 messy integration, I don't know how to do it, Wolfram says: http://integrals.wolfram.com/index.jsp?expr=1%2F%28x%5E2%2B2x%2B5%29%5E2&random=false
*Monday, March 12, 2012 at 12:33am by Reiny*

**Calc**

First make an appropriate substitution and then use integration by parts to evaluate the indefinite integrals: ∫ sinx cosx e^(1-sinx) dx I was going to substitute u= 1 - sinx, but then i got du = sinxcosx - xdx, so im stuck. Thank you so much!!
*Tuesday, October 18, 2011 at 9:02pm by Erica*

**brief calc**

Calculate the total area of the region described. Do not count area beneath the x-axis as negative. Bounded by the curve y = square root of x the x-axis, and the lines x = 0 and x = 16 This is under Integrals, i don't know what i'm doing wrong, please help
*Wednesday, November 6, 2013 at 9:05pm by kyle*

**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*

**Calc**

Break it into two integrals, integral (sec^2(x)tan(x))dx and integral(-tan(x))dx. Both can be solved using u substitution.
*Wednesday, May 30, 2012 at 8:17pm by Pendergast*

**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*

**Calc**

sin^3/cosx = (sinx/cos)(sin^2x) = tanx sin^2 x then http://integrals.wolfram.com/index.jsp?expr=%28tan%28x%29%28sin%28x%29%5E2%29&random=false same as above
*Monday, October 1, 2012 at 10:16pm by Reiny*

**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*

**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*

**calc**

This is why you'd need to do exercises in integration. To give you a hint, ∫dv/(g-kv) = -log(g-kv)/k This could be inferred from standard integrals: ∫dx/(a+bx) = (1/b)log(a+bx)
*Tuesday, September 8, 2009 at 2:42am by MathMate*

**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 - 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*

**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*

**Calc**

∫f(x)dx over [a,b] = ∫(g(x)+5)dx over [a,b]. By Addition rule of integrals, ∫(g(x)+5)dx = ∫g(x)dx + ∫5dx. = ∫g(x)dx + 5(b-a)
*Sunday, January 2, 2011 at 7:15pm by Marth*

**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**

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*

**calc II**

Express the integrals as the sum of partial fractions and evaluate the integral: (integral of) (x^2)dx/(x-1)(x^2 +2x+1) My work: The above integral is equal to x^2dx/(x+1)^2 (A/x-1) + (B/x+1) + (Cx+D)/(x+1)^2 = x^2 A(x+1)^2 + B(x-1)(x+1) + (Cx+d)(x-1) = x^2 Ax^2 + 2Ax + A + Bx...
*Sunday, December 6, 2009 at 9:59pm by Jenna*

**calc**

separate it into three integrals 19 dx --> 19 x ---> )19*3 - 19*1 2x^3 dx --> (1/2)x^4--> .5(27-1) -20xdx--> -10x^2-->10(1-9)
*Monday, November 10, 2008 at 4:38pm by Damon*

**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*

**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: 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*

**CALC -- integrals!**

2. f(x)=e^x for 0<x<2. Imagine that you have graphed this function in the x-y plane. For each value of x in the domain, erect a square above the curve perpendicular to the x-axis and parallel to the y-z plane where one side runs from the x-axis to the curve. So the ...
*Tuesday, February 26, 2008 at 6:43pm by anonymous*

**calc volumes**

the basse of a solid S is the region enclosed by the graph of y=square root (ln x), the line x=e, and the x-axis. if the cross sections of S perpendicular to the x-axis are squares, then the volume of S is. how do i find the side of the squares. because i got square root(lnx) ...
*Tuesday, December 5, 2006 at 1:21am by david*

**physics**

If you did not have the integrals to derive the equation, I bet you were given an equation like this: Potential Energy of charge close to other charge = PE or U = k Q1 Q2 / R That is all I did with the integrals. That is where the 1/Rend and 1/Rbegin comes from. The work done ...
*Saturday, February 23, 2008 at 2:42pm by Damon*

**SDSU**

∫(a^2 - bx^2) dx As a test, I let a=2 and b=3 and ran it through the Wolfram integrator. Looks at the mess I got with nice numbers.... I have idea. Do you have a table of integrals? See if you can find your pattern. http://integrals.wolfram.com/index.jsp?expr=%284-3x%5E2...
*Tuesday, February 7, 2012 at 7:14pm by Reiny*

**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*

**double integrals**

Combine the following two integrals into one by sketching the region, then switching the order of integration. (sketch the region) im gonna use the S for integral sign..lol SS6ycos(x^3-3x)dxdy+SS6ycos(x^3-3x) And the first integration limits for x are between -1 and y, for y ...
*Wednesday, March 21, 2012 at 10:50pm by Mary*

**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 **

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*

**calc**

find the area between the x-axis and the graph of hte given function over the given interval: y = sqrt(9-x^2) over [-3,3] you need to do integration from -3 to 3. First you find the anti-derivative when you find the anti-derivative you plug in -3 to the anti-derivative and ...
*Wednesday, April 18, 2007 at 9:06pm by mikayla*

**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*

**pre-calc**

oops! don't know why I put AP Calc, it's actually pre-calc....
*Friday, September 18, 2009 at 8:52pm by muffy*

**ap calc**

OOPS, I meant Pre-Calc not ap Calc sorry
*Saturday, October 10, 2009 at 2:45pm by MUFFY*

**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 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**

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**

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*

**CALC -- integrals!**

well, we know the intensity is exponential in time. I = Io e^-kt the half life is 8 days I/Io = .5 = e^-8k ln .5 = -8k -.693147=-8k k = .0866 so I = Io e^-.0866 t now total over 8 days, call it D for dose D = int I dt D = int dt Io e^-.0866 t from t=0 to t = 8 D = (Io/-.0866...
*Monday, February 4, 2008 at 2:47pm by Damon*

**Calc - Evaluating Integrals**

1.) upper bound: pi lower bound: 0 8sinx / sqrt(5-4cosx) dx 2.) upper bound: sqrt 3 lower bound: 0 4x / sqrt(x^2 + 1) dx 3.) upper bound: 1 lower bound: -1 (dx) / (3x-4)
*Wednesday, March 23, 2011 at 11:13pm by Amy*

**calculus**

my interpretation: (e^-x - 1)/(e^-x+x^2) , you have mis-matched brackets even the usual reliable Wolfram had difficulties with that one http://integrals.wolfram.com/index.jsp?expr=%28e%5E-x+-+1%29%2F%28e%5E-x%2Bx%5E2%29&random=false the way you typed it, fixing the missing ...
*Wednesday, April 9, 2014 at 8:59pm by Reiny*

**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*

**integrals**

integral of (x+6)/5
*Thursday, March 3, 2011 at 11:44pm by john2*

**Help W/ Definite Integrals**

You are welcome
*Sunday, April 29, 2012 at 4:05pm by Reiny*

**integrals**

What is the problem?
*Sunday, May 26, 2013 at 11:27am by me*

**MATHS**

A CONE WAS CONSTRACTED FROM PLASTIC, HAVE A HEIGHT OF 20cm (PART C). THE LID IS MADE UP OF HEMISPHERE HAVE RADIUS 0F 5cm (PART A) AND PYRAMID HAVE VOLUME OF 81,667cm. THE LID FITS EXACTLY OVER THE CONE. A) CALC. THE VOL. OF PART A. B) CALC. THE VOL. OF PART C. C) CALC. THE ...
*Monday, August 9, 2010 at 8:16am by MTHOMBENI BAFANA*

**Studying for Calc**

I have a quiz on derivatives coming up and I really want to do well, but i'm not sure how to go about studying calc. i have no trouble studying for other subjects, but i'm doing poorly in calc and i need to make some changes in my study habits. any suggestions? i usually work ...
*Tuesday, October 7, 2008 at 4:22pm by Elizabeth*

**Calculus (Integrals)**

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

**Calculus (Integrals)**

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

**Calc**

Can someone help me with this calc problem? lim as x->0 (tan^3 (2x))/(x^3)
*Thursday, January 8, 2009 at 4:22am by Nick*

**integral**

Well, I can refer you to the answer: http://integrals.wolfram.com/index.jsp?expr=e^x^%281%2F3%29&random=false
*Sunday, July 31, 2011 at 3:24pm by Damon*

**Tan^2 Integrals **

I'm having a hard time understanding how to do Integrals involving tan^2. I have two problems: 1. Find the integral of (tan^2 y +1)dy 2. Find the integral of (7tan^2 u +15)du 1. My approach to it is to replace the tan^2 y portion of the problem with sec^2 y -1, but it doesn't ...
*Tuesday, December 10, 2013 at 9:43am by Anonymous*

**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*

**Calc**

I'm creating a "calculu for dummies" project book. I've already done most of it, I'm just stuck on these ones. Thank you! -What is the definition of a derivative? -What does the derivative of a function tell you (in english) ? Then I need an example of the power rule, quotient...
*Thursday, May 19, 2011 at 6:12pm by Jill*

**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*

**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*

**mathematics**

find value of tan44degrees using integrals
*Sunday, January 23, 2011 at 10:03pm by rajni*

**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*

**Diploma Maths**

Evaluate the definite integrals [-2,0,(e^2+1),x] + [0,1,2 cos(x),]
*Friday, November 2, 2012 at 5:01pm by sean de la Harpe*

**Pre Calc**

Does this: sqrt 3x-4=(x-4)^2-3 Multiply out to: x^4-16x^3+90^2-211x+173 I then need to put it in the graphing calc and see what the roots are.
*Thursday, September 10, 2009 at 8:34pm by muffy*

**Pre-Calc/Calc**

Looks like C to me
*Wednesday, April 17, 2013 at 11:05am by Steve*

**calculus (limits)**

lim h>0 sqrt(1+h)-1/h not sure how to factor this; not allowed to use L'Hopital's Rule. (that isn't taught at my school until Calc II & I'm in Calc I).
*Saturday, January 22, 2011 at 2:09pm by John*

**Math**

Calculate the area above y=x^2 and to the right of x=y^2 using integrals.
*Wednesday, January 2, 2008 at 3:50pm by Bjorn*

**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**

Evaluate the given indefinite integrals. ∫ (x+2)(2x+3)^1/2
*Saturday, January 14, 2012 at 12:11am by Jack*

**Definite Integrals**

Integral [0,pi/2] of sin(x)ln(sinx) w.r.to x.
*Tuesday, June 26, 2012 at 9:25pm by alsa*

**Integrals**

Evaluate the integral of ∫(2x^2 −3)^2 dx
*Monday, May 20, 2013 at 8:31am by Grant*

**CALC -- integrals!**

Iodine-131 is used to treat hypo-thyroidism since it is preferentially absorbed by the thyroid and typically involves a total radiation dosage of 10,000,000 millirem. Iodine-131 has a half-life of eight days. 1. Set up the appropriate integral to represent the radiation dosage...
*Monday, February 4, 2008 at 2:47pm by anonymous*

**math**

Try evaluating it as 2 integrals by defining |x| piecewise. |x| = {x, x>=0; -x, x<0}
*Sunday, January 10, 2010 at 10:41am by Marth*

**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*

**Integrals- Log/Ln**

I'm still getting it wrong, so I'm not sure what's with that.. Thank you for your help though!
*Friday, May 24, 2013 at 12:52am by Erika*

**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*

**Math**

Looks like a rather unreasonable integral. Here is what Wolfram says http://integrals.wolfram.com/index.jsp?expr=x%5E3%2F%281%2Bx%5E6%29&random=false You also appear to have a typo for the upper end of your integral
*Saturday, April 14, 2012 at 8:52am by Reiny*

**Calculus - Integrals**

Explained here: http://www.jiskha.com/display.cgi?id=1206288758 If you have trouble understanding this, then the best thing you can do is to replacethe problem by simpler problems and see if you can solve them. E.g. try to integrate the function: 1/[(1+x)x^3] Assuming that you...
*Monday, March 24, 2008 at 12:53pm by Count Iblis*

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