Sunday

April 20, 2014

April 20, 2014

Number of results: 2,749

**Calculus - Integrals**

1) use the formula: cos(2x) = 2 cos^2(x) - 1 You can use this formula to derive a formula for cos(1/2 x) in terms of cos(x). You can then simplify the square root using that formula. 2) sec^4 (x) = 1/cos^4(x) You can use a reduction formula, which is easier to derive for ...
*Friday, March 14, 2008 at 9:00pm by Count Iblis*

**Calculus 3**

The integral over x is then from 0 to 1, the integral over y is from 0 to sqrt(1-x^2), the integral over z is from 0 to sqrt(1-y^2). You integrate the volume element dxdydz. Affter integrating over z, you are left with the integral over x and y of sqrt(1-y^2) dydx. Integrating...
*Friday, April 5, 2013 at 11:02am by Count Iblis*

**calculus**

You can proceed as follows. Substitute: x = e^t. Then the integral becomes: Integral from ln(2) to infinity of e^(t/2)/t dt We can get rid of the factor 2 in the exponential by putting y = t/2. The integral becomes: Integral from 1/2 ln(2) to infinity of e^(y)/y dy For ...
*Tuesday, August 25, 2009 at 7:52pm by Count Iblis*

**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*

**Calculus - integration**

Your expression came out with a weird symbol on my computer. Is that supposed to be an integral sign? Did you want [integral] (15^x/3^x) dx ?? 15^x/3^x = (15/3)^x = 5^x since you don't have any upper or lower limits on the integral, you want the indefinite integral [integral] ...
*Monday, November 1, 2010 at 10:39pm by Reiny*

**Integral**

That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten integral of sin^2 x = -sinx cos x + integral of (...
*Tuesday, February 20, 2007 at 5:07am by drwls*

**Calculus**

Leibnitz's Rule explains how to take the derivative of an integral. Take a google for it, or consult your textbook. Basically, you have a product here. cos(x) * Integral f(x) d/dx of the product is -sin(x) * Integral + cos(x) * d/dx(Integral(f)) d/dx(Integral) = Integral(df/dx...
*Monday, October 17, 2011 at 4:18am by Steve*

**Calculus 3**

The integral of sin(x)sin(y)dxdy over the region is... (it is easy, the integral factorizes into two one dimensional integrals. Then you must divide this integral by...(the integral is the average times... )
*Friday, April 15, 2011 at 1:34pm by Count Iblis*

**double integral**

1. Sketch the region of integration & reverse the order of integration. Double integral y dydz... 1st (top=1, bottom =0)... 2nd(inner) integral (top=cos(piex), bottom=(x-2)... 2. Evaluate the integral by reversing the order of integration. double integral sqrt(2+x^3) dxdy... ...
*Friday, March 11, 2011 at 12:13am by Michelle*

**calculus**

sin(x) + cos(x) = sqrt(2) sin(x + pi/4) So, the integral can be written as: pi/4 ln(2) + Integral of Log[sin(x)]dx from 0 to pi/2 Let's call the integral in here I: I = Integral of Log[sin(x)]dx from 0 to pi/2 Then note that because sin is symmetric w.r.t. reflection about x...
*Monday, June 11, 2012 at 3:21pm by Count Iblis*

**college math**

You only have to do the ln thing if it is the integral of 1/x otherwise integral x^n dx = (1/n+1) x^(n+1) that works fine for n = -2 integral x^-2 dx = (1/-1) x^-1 as I said HOWEVER in the nasty case of integral x^-1 dx = (1/0) x^0 BUT 1/0 is UNDEFINED
*Sunday, February 15, 2009 at 7:34pm by Damon*

**Calculus II/III**

A. Find the integral of the following function. Integral of (x√(x+1)) dx. B. Set up and evaluate the integral of (2√x) for the area of the surface generated by revolving the curve about the x-axis from 4 to 9. For part B of our question , the surface of revolution ...
*Monday, February 19, 2007 at 2:00pm by Ryoma*

**Calculus**

Find the volume of the solid whose base is the region in the xy-plane bounded by the given curves and whose cross-sections perpendicular to the x-axis are (a) squares, (b) semicircles, and (c) equilateral triangles. for y=x^2, x=0, and y=0 (a) integral (x^2)^2 from 0 to 2=32/5...
*Saturday, December 15, 2007 at 11:41pm by Anonymous*

**Calculus**

You can substitute x = sqrt(2/k) y. That yields: I = integral over (0,+∞) of e^(-k/2*x^2) dx = sqrt(2/k) integral over (0,+∞) of e^(-y^2) dy Then, the result follows from the fact that integral over (0,+∞) of e^(-y^2) dy = 1/2 sqrt(pi) To prove that, you can ...
*Tuesday, August 25, 2009 at 6:17pm by Count Iblis*

**HELP WITH MATH**

This looks like integral calculus, so why do they call it precalculus? The area is the definite integral of the function y(x) from 3 to 6. The definite integral is the change in the indefinite integral going from x=3 to x=6. The indefinite integral is (2/3)x^3 +x^2/2 + x ...
*Wednesday, May 14, 2008 at 9:47pm by drwls*

**Math/Calculus**

How would I solve the following integral with the substitution rule? Integral of: [(x^3)*(1-x^4)^5]dx Put 1-x^4 = y Then -4x^3 dx = dy Integral is then becomes: Integral of -1/4 y^5 dy ok, thanks a lot! I got it now.
*Monday, May 28, 2007 at 9:33pm by COFFEE*

**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*

**Integral Help**

I need to find the integral of (sin x)/ cos^3 x I let u= cos x, then got -du= sin x (Is this right correct?) I then rewrote the integral as the integral of -du/ u^3 and then rewrote that as the integral of - du(u^-3). For this part, I wasn't sure how to finish. I was hoping to...
*Thursday, November 14, 2013 at 12:44am by Anonymous*

**calculus**

consider the function f(x) = e^x(sinNx) on the interval [0,1] where N is a positive integer. a) Compute the integral from 0 to 1 of f(x). Evaluate this integral when N=5, N=10, and N=100. B) What happens to the graph and to the value of the integral as N-->infinity? Does ...
*Tuesday, February 23, 2010 at 4:33pm by Lindsay*

**calculus**

consider the function f(x) = e^x(sinNx) on the interval [0,1] where N is a positive integer. a) Compute the integral from 0 to 1 of f(x). Evaluate this integral when N=5, N=10, and N=100. B) What happens to the graph and to the value of the integral as N-->infinity? Does ...
*Tuesday, February 23, 2010 at 7:29pm by anonymous*

**Calculus AP**

The integrand is an odd function of x, therefore the integral will be zero. As an exercise try to prove that if f(-x) = -f(x) that the integral from -r to r of f(x) dx is zero, e.g. by splitting it from -r to zero and from zero to r and then by substituting x =-t in the first ...
*Sunday, August 26, 2012 at 11:00pm by Count Iblis*

**calc asap!**

can you help me get started on this integral by parts? 4 S sqrt(t) ln(t) dt 1 please help! thanks! Integral t^(1/2)Ln(t)dt = 2/3 t^(3/2)Ln(t)- 2/3 Integral t^(1/2) dt = 2/3 t^(3/2)Ln(t) - 4/9 t^(3/2) Simpler method: Integral t^(a)dt = t^(a+1)/(a+1) Integral d/da [t^(a)]dt = d/...
*Thursday, June 7, 2007 at 6:30pm by Woof*

**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*

**Math**

Identify u and du for the integral. 1. The integral of [(cosx)/(sin^(2)x)]dx 2. The integral of sec2xtan2xdx
*Tuesday, November 30, 2010 at 9:07pm by Jess*

**math**

evaluate the double integral and reverse order of integration [(first integral 0 to 1)(second integral 9y to 9)e^(x^2)dx)dy
*Sunday, May 9, 2010 at 5:25pm by college*

**Calculus 2**

The question is: Evaluate the improper integral for a>0. The integral is: the integral from 0 to infinity, of e^(-y/a)dy Can anyone help me solve this? When I try I get 'a', which apparently is incorrect. Thank you!
*Sunday, September 25, 2011 at 4:10pm by Sara*

**Math**

1. For the indefinite integral, try the substitution ln x = u e^u = x e^u du = dx indef. integral of dx/(x(1+ln(x)) = integral of e^u du/[e^u*(1 + u) = integral of du/(1+u) = ln (1 + u) = ln (1 + ln x) When x = e^6, the indef. integral is ln (1 + 6) = ln 7 = 1.9459 When x = 1...
*Friday, January 30, 2009 at 2:37am by drwls*

**Calculus**

Integrate the function over that interval and divide the integral by e^2 - 1. Try integration by parts for the integral. Let lnx = u and dv = 6 x dx du = dx/x and v = 3 x^2 Integral of f(x) dx =Integral u dv = uv - Integral v du = 3 x^2 lnx - Integral of 3x dx = 3x^2 lnx - (3/...
*Saturday, November 26, 2011 at 5:51pm by drwls*

**calc**

Try integration by parts. Let u = ln x du = dx/x dv = 1/x^2 dx v = -1/x Integral of lnx/x^2 = Integral of udv = uv - Integral of vdu = -(ln x)/x - Integral of -dx/x^2 Take it from there, but check my work also
*Wednesday, January 23, 2008 at 10:36pm by drwls*

**calculus**

If f(x) is odd, then f(-x) = -f(x) So, integral from -3 to 0 is minus the integral from 0 to 3 So, what's left is just integral from 3 to 7 = 11.
*Wednesday, November 30, 2011 at 6:21pm by Anonymous*

**Calc BC**

1. Find the indefinite integral. Indefinite integral tan^3(pix/7)sec^2(pix/7)dx 2. Find the indefinite integral by making the substitution x=3tan(theta). Indefinite integral x*sqrt(9+x^2)dx 3. Find the indefinite integral. Indefinite integral cos(x)sin^3(x)dx
*Tuesday, September 27, 2011 at 7:24pm by Zooey*

**Calculus**

evaluate the integral or state that it diverges. Check if I did it correctly. integral 0,1 dr/r^.999 lim b->0+ integral b, 1 1000r^.001 =-1000
*Friday, February 8, 2008 at 11:52pm by Anonymous*

**Calculus**

Introduce polar coordinates in the y-z plane. You can then write the integral as: Integral over theta from 0 to 2 pi Integral over r from 0 to 1 Integral over x from 10 r^2 to 10 r dr dtheta dx
*Friday, April 23, 2010 at 7:34pm by Count Iblis*

**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*

**calc**

how do you start this problem: integral of xe^(-2x) There are two ways: 1) Integration by parts. 2) Differentiation w.r.t. a suitably chosen parameter. Lets do 1) first. This is the "standard method", but it is often more tedious than 2) You first write the integral as: ...
*Tuesday, May 15, 2007 at 7:55pm by walex*

**calculus (integral)**

evaluate the integral: integral of 13/(x^3 - 27) dx
*Friday, September 13, 2013 at 2:45am by Anonymous*

**calculus**

Evaluate the integral of 1/x^3 from x = 1 to x = 2. That is the enclosed area of the region. I get 3/8. See what you get. Then pick the upper limit of integration, b, such that the integral of 1/x^3 from 1 to b is half the number you got for the full integral.
*Tuesday, January 26, 2010 at 4:44am by drwls*

**calculus**

Is f(x) the probability of waiting x? If so, the mean time is Integral x* f(x) dx/Integral f(x) dx Both integrals go from 0 to infinity. The denominator integral is 1. mean waiting time = Integral x* f(x) dx = Integral (x/8) e^(-x/8) dx = Integral 8u e^-u du where u = x/8 and ...
*Monday, April 7, 2008 at 10:04pm by drwls*

**calculus**

1. integral -oo, oo [(2x)/(x^2+1)^2] dx 2. integral 0, pi/2 cot(theta) d(theta) (a) state why the integral is improper or involves improper integral (b) determine whether the integral converges or diverges converges? (c) evaluate the integral if it converges CONFUSE: how would...
*Sunday, February 10, 2008 at 1:51am by Anonymous*

**calculus **

LEt f and g be continous functions with the following properties i. g(x) = A-f(x) where A is a constant ii. for the integral of 1 to 2 f(x)dx= the integral of 2 to 3 of g(x)dx iii. for the integral from 2 to 3 f(x)dx = -3A a find the integral from 1 to 3 of f(x)dx in terms of ...
*Sunday, February 13, 2011 at 5:01pm by Little*

**calculus **

LEt f and g be continous functions with the following properties i. g(x) = A-f(x) where A is a constant ii. for the integral of 1 to 2 f(x)dx= the integral of 2 to 3 of g(x)dx iii. for the integral from 2 to 3 f(x)dx = -3A a find the integral from 1 to 3 of f(x)dx in terms of ...
*Sunday, February 13, 2011 at 6:41pm by alex *

**math**

LEt f and g be continous functions with the following properties i. g(x) = A-f(x) where A is a constant ii. for the integral of 1 to 2 f(x)dx= the integral of 2 to 3 of g(x)dx iii. for the integral from 2 to 3 f(x)dx = -3A a find the integral from 1 to 3 of f(x)dx in terms of...
*Monday, February 14, 2011 at 6:21pm by jimbo*

**Calculus integral**

evaluate the integral: integral from -pi/4 to 0 for the function 6sec^3x dx. it has to be an exact answer and i did it and keep getting it wrong. I got 4sqrt(2)-4ln(-sqrt(2)+1)
*Tuesday, October 16, 2012 at 12:40am by kajri*

**calculus**

8). Part 1 of 2: In the solid the base is a circle x^2+y^2=16 and the cross-section perpendicular to the y-axis is a square. Set up a definite integral expressing the volume of the solid. Answer choices: integral from -4 to 4 of 4(16-y^2)dy, integral from -4 to 4 of (16+y^2)dy...
*Sunday, April 14, 2013 at 9:19pm by Sally*

**calculus 2**

Let X = a + bx + c x^2, with a = sqrt2, b = 0 and c = 1. The integral of x^5/X is x^4/4 - sqrt2*(Integral of x^3 dx/X) Use the recursion formula again for the second term. The integral of x^3/X is x^2/2 - sqrt2*(Integral of x dx/X) The Integral of x dx/X is (1/2)ln X = (1/2)ln...
*Wednesday, March 30, 2011 at 12:27am by drwls*

**Brief Calculus**

2*x^2=t 2*2xdx=dt 4xdx=dt Divide with 4 xdx=dt/4 Integral of x/(2x^2−1)^0.4 dx= Integral of dt/4(t-1)^0.4= (1/4) Integral of (t-1)^(-0.4)dt Integral of x^n=x^(n+1)/(n+1) Integral of (t-1)^(-0.4)dt= (t-1)^(-0.4+1)/(-0.4+1)+C= (t-1)^0.6/0.6+C=(t-1)^(3/5)/0.6+C Integral of ...
*Monday, May 2, 2011 at 4:57am by Anonymous*

**calculus**

displacement = integral v dt = -t^3/3 + 2 t^2 - 3 t from -2 to +6 do the arithmetic distance traveled that means always use absolute value of any part of the integral when is velocity negative and when is it positive? 0 = t^2 - 4 t + 3 0 = (t-3)(t-1) velocity 0 at t = 1 and t...
*Friday, December 9, 2011 at 4:31am by Damon*

**functions**

That equals the ratio of the integral of e^-x^2 from 0 to 2 to the integral of e^-x^2 from 0 to infinity. (The curve is symmetric about the y axis, so you only need to do one half of the integration.) You can find the answer in tables of the probability integral. It is 0.99532
*Monday, November 24, 2008 at 11:59pm by drwls*

**calculus**

2*x^2=t 2*2xdx=dt 4xdx=dt Divide with 4 xdx=dt/4 Integral of x/(2x^2−1)^0.4 dx= Integral of dt/4(t-1)^0.4= (1/4) Integral of (t-1)^(-0.4)dt Integral of x^n=x^(n+1)/(n+1) Integral of (t-1)^(-0.4)dt= (t-1)^(-0.4+1)/(-0.4+1)+C= (t-1)^0.6/0.6+C=(t-1)^(3/5)/0.6+C Integral of ...
*Monday, May 2, 2011 at 4:50am by Anonymous*

**Calculus**

That is a fine substitution to make u = x + 1 dx = du x^2 = u^2 - 2u+1 so we have integral (u^2-2u+1)(u^.5) du that is integral of u^2.5 du - integral of 2u^1.5 du + integral of u^.5 du in general integral of u^n du = [u^(n+1)]/(n+1)
*Sunday, December 9, 2007 at 3:26pm by Damon*

**calc**

is y = 2x - x^2 the function given? if it is, area under the curve means the integral at these particular bounds,, first you have to recall how to integrate: for ax^n (a is constant), add 1 to the power, then this n+1 must be divided: integral (ax^n) = (a)[x^(n+1)]/(n+1) thus ...
*Thursday, October 28, 2010 at 11:05am by jai*

**CALCULUS:)**

(e^x+1)/(e^x +x) = f'(x)/f(x) with f(x) = e^x +x Integral of f'(x)/f(x) dx can be computed by putting f(x) = y then f'(x)dx = dy So, the integral can be written as: Integral of dy/y Integration limits: if x = 0, then y = f(x) = 1 if x = 1, then y = f(1) = 1 + exp(1) So, the ...
*Tuesday, August 9, 2011 at 8:11pm by Count Iblis*

**math**

Evaluate the following indefinite integral by using the given substitution to reduce the integral to standard form integral 2(2x+6)^5 dx, u=2x+6
*Tuesday, September 10, 2013 at 12:13am by vishal*

**math**

Linearly Independent Vector u2 = e^x Orthogonal Vector v2 = e^x - e + 1 |v2|^2 = Integral from zero to 1 of [e^x - e + 1]^2 dx = Integral from zero to 1 of [e^(2x) + 2 (1-e)e^(x) + (1-e)^2] dx = 1/2 (e^2 -1) - (e-1)^2 = (e-1)[1/2 (e+1) - (e-1)] = 1/2(e-1)(3 - e) So, g2 should ...
*Friday, January 18, 2008 at 12:13am by Count Iblis*

**math**

The integral of what? you have 2 different functions and their graphs Are you finding the area between? the integral of sinx is -cosx and the integral of tanx = -ln(cosx) Does that get you anywhere?
*Wednesday, November 28, 2012 at 9:48am by Reiny*

**math**

Evaluate the following indefinite integral by using the given substitution to reduce the integral to standard form integral cos(9x) dx, u=9x
*Tuesday, September 10, 2013 at 12:14am by vishal*

**Calculus**

Substituting x = sqrt(t) leads to an integral of the form: Integral of dt/t sqrt(1+t^2) If you then put t = sinh(u), this becomes: Integral of cosh^2(u)/sinh(u)du = Integral of [1/sinh(u) + sinh(u)] du Then the integral of 1/sinh(u) be evaluated by putting u = Log(v): du/[exp(...
*Tuesday, July 3, 2012 at 6:03pm by Count Iblis*

**Calculus 3**

Write it as the repeated integral: Integral over y from 0 to 1 dy Integral over x from 0 to y dx e^(y^2) = Integral over y from 0 to 1 dy ye^(y^2) = 1/2 [exp(1) - 1]
*Monday, December 10, 2007 at 6:38pm by Count Iblis*

**Integral calculus**

You have not proved to me that the Integral of lnsinx w.r.t.x from 0 to pi/2 is equal to:Integral of lnsinx w.r.t.x from 0 to pi/4 plus Integral of lncosx w.r.t.x from 0 to pi/4.
*Wednesday, June 13, 2012 at 5:51pm by alsa*

**Calculus**

let x^(1/3) = w w^3 = x dx = 3w^2 dw The integral becomes 3 *INTEGRAL w^2 sin w dw The integral of that is in my table of integrals and it looks like you need to apply integration by parts twice to get it.
*Friday, February 13, 2009 at 12:23am by drwls*

**Calculus: Integral**

I don't understand how to do this one integral problem that involves secant. I'm asked to find the integral of sec^4 (4x). I'm not really sure how to go about solving this problem.
*Monday, March 31, 2014 at 3:32am by Anonymous*

**Calculus**

Integral of x / sqrt(x-1) dx We can use substitution method. Let u = x - 1 Thus, x = u + 1 and dx = du Substituting, Integral of (u+1)/sqrt(u) du Integral of (u+1)(u^(-1/2)) du Integral of u^(1/2) + u^(-1/2) du = (2/3)*u^(3/2) + (2)*u^(1/2) + C = (2/3)*(x-1)^(3/2) + 2(x-1)^(1/...
*Wednesday, November 27, 2013 at 7:22am by Jai*

**Calc II**

Perform long division on the integrand, write the proper fraction as a sum of partial fractions, and evaluate the integral: (integral of) 2y^4dy/y^3 - y^2 + y - 1 After long divison I get: (integral of)2ydy + 2(integral of)dy + (integral of) 2/y^3 - y^2 + y - 1 I keep getting ...
*Sunday, December 6, 2009 at 10:08pm by Jenna*

**calc 2**

Determine whether the integral converges or diverges. Find the value of the integral if it converges. The integral where b=2 and a=0 (x/x^2-1 dx).
*Thursday, September 23, 2010 at 11:53am by Pete*

**Calc or Pre calc**

To get the definite integral subtract the value of the indefinite integral at x=2 from the value at x=5. The arbitrary cnstant c will cancel out. The indefinite integral is x^4/4 - x^2 + C. The definite integral is (625/4) - 25 -[16/4 - 4].
*Wednesday, May 14, 2008 at 9:42pm by drwls*

**Calculus**

Can someone look over my work and tell me if my steps look correct? I'm trying to correct some problems that looked wrong. Instructions: Find the total areas between the given curves. 1. x= (y^3) and x=(y^2) on the interval [0,1] (integral from 0 to 1 of) ((y^3)-(y^2))dy = (...
*Thursday, September 10, 2009 at 7:47pm by Jenna*

**Math**

The indefinite integral of dv/(z^2-v^2), with z being a constant, is [1/(2z)]log[(z+v)/(z-v)] Evaluate that at v=v' and subtract the value for v=0, to get the definite integral. The method of partial fractions can used to get the integral. It involves rewriting 1/[(z^2-v^2} as...
*Thursday, December 6, 2007 at 1:05am by drwls*

**Integration-Calculus**

is it just integral 1 + integral f(x) and the integral 1 = x?
*Sunday, February 14, 2010 at 2:18am by so CONFUSED!*

**Calculus**

But how did you get Integral of x sq/(1+x sq) dx = integral (1/2)u^(1/2) du du= 2xdx, from there i got, integral xsq/(1+x) dx = (1/2) integral[ (sqrt (1-u))/u ]du Did i do anything wrong?
*Monday, December 7, 2009 at 1:36pm by O_o Rion*

**definite integral**

Use the Riemann Sums corresponding to 5 inscribed rectangles of equal width to approximate the integral a= 1, b= 3, (1/x)dx this is all for definite integral i just know x1=1.4, x2=1.8, x3=2.2, x4=2.6, x5=3.0 how do i continue
*Wednesday, May 15, 2013 at 10:51am by Eric*

**Calculus**

If w = x^5, 5x^4 dx = dw The integral you want becomes the integral of (1/5) w cos w dw Now use integration by parts, with u = w dv = cosw dw du = dw v = sin w The integral becomes (1/5)[u v - Integral v du] = (1/5)[w sin w - integral of sin w dw] = (1/5)[w sin w + cos w] = (1...
*Sunday, February 22, 2009 at 9:37pm by drwls*

**Math/Calculus**

How would I evaluate the following integral by using integration by parts? Integral of: (t^3)(e^x)? You mean (x^3)(e^x)? x^3 exp(x) dx = x^3 d[exp(x)] = d[x^3 exp(x)] - exp(x) d[x^3] = d[x^3 exp(x)] - 3 x^2 exp(x) dx So, if you integrate this you get x^3 exp(x) - 3 Integral of...
*Monday, May 28, 2007 at 9:38pm by COFFEE*

**Calc**

Evaluate the integral using any method: (Integral)sec^3x/tanx dx I started it out and got secx(1tan^2x)/tanx. I know I just have to continue simplifying and finding the integral, but I'm stuck on the next couple of steps. Also, I have another question witht he same directions...
*Monday, October 1, 2012 at 10:16pm by H*

**calculus**

Write f(x) as f(x) = x - 1/x, and integrate it one term at a time. The integral (or "antiderivative", if you insist on using that term) of x is (x^2)/2. The integral of -1/x is -ln x. An arbitrary constant term can be added to the integral.
*Sunday, November 8, 2009 at 7:07pm by drwls*

**Math**

integral | 2 x | dx = 2 integral | x | dx = 2 ( x ^ 2 ) * sgn ( x ) + C = integral | 2 x | dx from - k to k = k ^ 2 - [ ( k ^ 2 ) * ( - 1 ) ] = k ^ 2 + k ^ 2 = 2 k ^ 2 2 k ^ 2 = 18 Divide both sides by 2 k ^ 2 = 9 k = sqrt ( 9 ) = ± 3 integral | 2 x | dx from - 3 to 3 = 18 k = 3
*Saturday, April 14, 2012 at 5:12am by Bosnian*

**Calc**

If we know that the definite integral from -6 to -3 of f(x) equals 6, the definite integral from -6 to -5 equals 2 and the definite integral from -4 to -3 equals 4 then: What is the definite integral from -5 to -4? I know that this is zero. But then what can we say about the ...
*Friday, September 27, 2013 at 8:07pm by Anonymous*

**Physics (Mechanics)**

The half-width of the parabolic section at any height y above the x axis is x = sqrt (y/1.1) Perform one integral for the total mass in terms of H, using the known mass. This will tell you the density x thickness. Then perform a second integral for the moment of inertia, using...
*Friday, March 21, 2008 at 10:04pm by drwls*

**integral**

The answer is the integral of F dx from x=1 to 4. That would equal the value of the indefinite integral x^3 -x^2/2 + x at x=4 MINUS the value at x+1. [64 -8 + 4] - [1 - 1/2 + 1] = 60 - (3/2) = 117/2 Check both our calculations; I am getting sloppy and have posted several wrong...
*Friday, August 7, 2009 at 3:20pm by drwls*

**intergrals**

find value of def integral with a=-2 and b=2sqrt(3) definite integral is : x^3 * sqrt(x^2+4) dx for integral i get 1/15 *((4+x^2)^(3/2)) (-8+3x^2) for value i get [1536- 64sqrt(2)]/15 but its' wrong. help please
*Sunday, May 26, 2013 at 4:36pm by Mikey*

**Calculus Theorem! !!!????**

There is a fundamental theorem oc calculaus that states that the integral of f(x)dx from a to b is G(b) - G(a). This assumes than f is the derivative of G (or G is the integral of f). Therefore G(4) = G(2) + integral(2 to 4)of f(x)dx What you wrote is not correct. The -7 ...
*Tuesday, March 23, 2010 at 12:55am by drwls*

**calc**

also: integral of tan^(-1)y dy how is integration of parts used in that? You write: arctan(y)dy = d[y arctan(y)] - y d[arctan(y)] Here we again have used the product rule: d(fg) = f dg + g df You then use that: d[arctan(y)] = 1/(1+y^2) dy So, the integral becomes: y arctan(y...
*Wednesday, May 23, 2007 at 5:34pm by marsha*

**math **

Can someone help me answer this? If a < 5 the define integral [a, 4] 2.4e^(1.4x)dx = 44 Find the value a Define integral = integral sign a = lower limit 4 = upper limit
*Sunday, May 19, 2013 at 8:05pm by mark *

**indefinite integrals**

The indefinite integral is x^(1/2)/(1/2) = 2 x*^(1/2) Evaluate that at x=0 and subtract the value ax x=1. That equals -2. The integral from 0 to 1 is +2. The integral remains finite in spite of the infinite value of the integrand at x=0.
*Thursday, August 27, 2009 at 12:45am by drwls*

**Calculus**

F(x) = cos(x) • the integral from 2 to x² + 1 of e^(u² +5)du Find F'(x). When i did this, i got: -2xsin(x)e^((x²+1)² + 5) But my teacher got: -sin(x) • the integral from x² + 1 of e^(u² +5)du + 2xcos(x)e^((x²+1)² + 5) Do you know why the integral is in his answer? I'm not sure...
*Monday, October 17, 2011 at 4:18am by Erica*

**Calculus**

Which of the following is a step in evaluating. (Integral) cos^2 5x dx A. (Integral) 1+cos10x/2 dx B. (Integral) 1-cos10x/2 dx C. (Integral) 1+cos10x/20 dx D. (Integral) 1-cos10x/20 dx
*Saturday, August 4, 2012 at 9:36pm by Cynthia*

**Calculus**

Use the symmetry of the graphs of the sine and cosine functions as an aid in evaluating each definite integral. (a) Integral of sinx*dx from -pi/4 to pi/4 (b) Integral of cosx*dx from -pi/4 to pi/4 (c) Integral of cosx*dx from -pi/2 to pi/2 (d) Integral of sinx*cosx*dx from -...
*Sunday, November 15, 2009 at 9:19pm by John*

**Calculus**

Is your 9^1 a misprint? What is the purpose of the 1? Why is there no "dt" in your integral? Both of your answers are incorrect, since they are constants. The indefinite integral would be a function of t, and the definite integral would have to have specified limits of ...
*Tuesday, March 3, 2009 at 7:36pm by drwls*

**Math**

Find the integrals. (show steps) (integral sign) xe^(4x^2) I think this how is how its done: (integral sign) xe^(4x^2) it's a u du problem let u=4x^2 so, du=8x dx now you have an x already so all u need is 8 inside and and 1/8 outside the integral [1/8] (integral sign) [8]xe^(...
*Monday, December 11, 2006 at 12:40pm by Jay*

**math**

How do I derive the secant reduction rule? Integral (sec x)^n dx = Integral (sec x)^(n-2) * (sec x)^2 dx = Integral ((tan x)^2 + 1)^(n/2-1) * (sec x)^2 dx Doing a substitution with: u = tax x du = (sec x)^2 dx = Integral (u^2 + 1)^(n/2-1) * du At this point I'm stuck. Any ...
*Wednesday, November 7, 2007 at 8:57pm by mathstudent*

**Calc**

Why would you say that? I understand you use u subsitution u= x^3 and then eventually you get an integral from 0 2 of the function but i dnt understand why the integral from 0 to 2 and the integral from 0 to 8 would be the same?
*Tuesday, May 4, 2010 at 2:40pm by Lucy*

**Calculus**

Evaluate the Integral 1. integral of (x^9+7x^6-1)/x^8 dx 2. integral of x^(1/3)*(42-x)^2 dx 3. integral of 9x+5/7x^3 dx
*Sunday, July 25, 2010 at 3:55pm by Mely*

**calculus **

Evaluate the Integral 1. integral of (x^9+7x^6-1)/x^8 dx 2. integral of x^(1/3)*(42-x)^2 dx 3. integral of 9x+5/7x^3 dx
*Sunday, July 25, 2010 at 7:45pm by Mely*

**Calculus**

I need help with this integral. w= the integral from 0 to 5 24e^-6t cos(2t) dt. i found the the integration in the integral table. (e^ax/a^2 + b^2) (a cos bx + b sin bx) im having trouble finishing the problem from here.
*Sunday, August 30, 2009 at 9:19pm by Bobby Smith*

**calculus**

how do you solve the integral of 1/[(square root of x)(lnx)] from 2 to infinity? i did the p- integral theorem with 1/square root of x and got it to be a divergent integral. however i was told this was the wrong way and that i should do it by integration by parts. but i can't ...
*Tuesday, August 25, 2009 at 7:52pm by kathryn*

**Calc 121**

How do you integrate using substitution: the integral from 1 to 3 of: ((3x^2)+(2))/((x^3)+(2x)) There is a trick to this one that grealy simplifies the integral. Let u = x^3 + 2x. Then du = (3x^2 + 2)dx The integral then bemoces just the integral of du/u, which is ln u = ln (x...
*Saturday, April 21, 2007 at 10:05pm by Me*

**AP Calculus**

you will get the same solid if you rotate y = x^2 + 1 about the x-axis. then volume = [integral] pi(y^2)dx from -1 to 1 or = 2pi[integral](x^2+1)^2 dx from 0 to 1 = 2pi[integral](x^4 + 2x^2 + 1) dx from 0 to 1 = ... this is quite easy to integrate and finish.
*Sunday, March 22, 2009 at 6:04pm by Reiny*

**math**

Generalize this to fine a formula for the integral: sin(ax)cos(bx)dx Could someone tell me what they got for an answer so I can check it to see if my answer is right. My answer: -1/2sinasinbx^2-1/3acosaxcosbx^3+ integral 1/3 a^2cosbx^3sinax..I'm not sure hot to find the ...
*Monday, February 25, 2008 at 3:50am by Jessica*

**math **

Sorry I posted this question earlier with a error. Can anyone help me solve this now? Can someone help me answer this? If a < 5 the define integral [a, 4] 2.4e^(1.4x)dx = 44 Find the value a Define integral = integral sign a = lower limit 5 = upper limit
*Sunday, May 19, 2013 at 8:38pm by mark *

**calculus**

yes. the substitution is correct: let u = x^(1/2) thus du = 1/[2(x^(1/2))] dx, or dx = 2(x^(1/2)) du, or dx = 2u du substituting these to original integral, integral of [e^x^(1/2) / x^(1/2)] dx integral of [(e^u) / u] * (2u) du the u's will cancel out: integral of [2*e^u] du ...
*Monday, May 2, 2011 at 10:41pm by Jai*

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