Number of results: 2,620
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
Please can anyone help with the following problems - thanks. 1) Integrate X^4 e^x dx 2) Integrate Cos^5(x) dx 3) Integrate Cos^n(x) dx 4) Integrate e^(ax)Sinbx dx 5) Integrate 5xCos3x dx The standard way to solve most of these integrals is using partial integration. So, look ...
Friday, April 13, 2007 at 9:05am by Febby
integral of Sec[2x]Tan[2x] i know u is sec 2x du=2sec2xtan2x dx what would i have to multiply with du so it would equal tan 2x dx? if my question is confusing, then here's another example of what i'm talking about: integral of (3x-2)^30 dx u=3x-2 du=3 dx 1/3 du=dx (i need help...
Monday, November 26, 2007 at 11:21pm by julie
What is the integral of 1/(x√(x^2-4)) dx? So I know that I am going to have to factor out the 4 and use the arcsec trig identity but I've having trouble getting to a viable u-substitution. I've gotten to factoring like this so far: √(x^2-4) = √[2^2(x^2/2^2 - ...
Wednesday, September 3, 2008 at 7:10am by Terry
I'm extremely rusty at these, but I think it's possible you might be able to use the substitution x=2cosh(u), where cosh is the hyperbolic cosine. cosh²(u)-sinh²(u)=1, so sqrt(x²-4) = sqrt(4cosh²(u)-4) = 2sinh(u). dx/du=2sinh(u), so dx=2sinh(u)du, which cancels out with the ...
Wednesday, September 3, 2008 at 7:10am by David Q
not sure what you mean by "area length" 1. do you want the area of the parabola from x = 0 to x = 2, which is where the parabola cuts the x-axis? If so, then it would simply be Integral (2x - x^2)dx from 0 to 2 = (x^2 - 1/3(x^3)│from 0 to 2 = 4 - 8/3 = 4/3 2. Do you want...
Monday, September 15, 2008 at 10:45am by Reiny
You need x as a function of y to do the integration. In this case, x(y) = 1/(8y^2) + (y^4)/4 To get the surface area, evaluate: Integral of 2 pi x(y) dy (y = 1 to 2) The indefinite integral is 2 pi [-1/(24y^3) + y^5/20] = pi[(y^5)/10 - 1/(24y^3)] Evaluate it at y=2 and y=1, ...
Sunday, September 21, 2008 at 4:30am by drwls
find the following integral (1) dx/25+(x-5)^2 (2) cos^8xsinxdx
Thursday, August 6, 2009 at 1:46pm by matrix school 2
(1) Do you mean dx/(25+(x-5)^2) or dx(1/25+(x-5)^2)? (2) (cos(x))^8sin(x)dx let u = cos(x) du = - sin(x) = integral(-u^8du) = -(u^9)/9 +C = -((cos(x))^9)/9 +C
Thursday, August 6, 2009 at 1:46pm by Marth
find the integral ~(e^x + x)^2 (e^x + 1)dx can you pls explain how it arrive to this? ans = 2e^4x + C
Thursday, August 6, 2009 at 3:05pm by matrix school 2
find the following integral dx/(25+(x-5)^2) yes this.
Thursday, August 6, 2009 at 2:11pm by matrix school 2
The integral(u' / (a^2 - u^2)) = 1/a(atan(u/a)) +C let u = x-5 du = dx and let a = 5 So the answer is 1/5 (atan((x-5)/5)) +C Note: atan = arctangent
Thursday, August 6, 2009 at 2:11pm by Marth
integral( (e^x+ x)^2 * (e^x + 1) dx ) let u = e^x + x du = (e^x + 1)dx Now you have integral(u^2 du) = (u^3)/3 +C = ((e^x + x)^3)/3 +C = (e^3x + 3x*e^2x + 3x^2*e^x + x^3)/3 +C Are you looking at the right answer key?
Thursday, August 6, 2009 at 3:05pm by Marth
From your description, the enclosed region is an area, not a volume. It is simply the integral of x^2 from 0 to 5. Did you leave out a part about the area being rotated about the x axis? That would give you a paraboloidal volume
Friday, August 7, 2009 at 7:29am by drwls
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
can any one explain how to evaluate this improper integral i.e. the function is not continuous at 0 neither at inf integration of [(e^(-sqrt(t)))/sqrt(t)]dt from 0 to infinity the integration part is easy but i want only how to evaluate it thanks
Sunday, May 2, 2010 at 2:07am by William
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
Maths Calculus Derivative Integral - Urgent Please
Use 2nd Fundamental Theorem of Calculus to find derivative of f(x) = integral of 2x^2 (at the top) to x-5 (at the bottom) of square root of Sin(x)dx
Monday, October 18, 2010 at 8:26am by Nick
Maths Calculus Derivative Integral - Urgent Please
Hmmmm. You want the derivative of the integral. well the integral is -cos(2x^2)+cos(x-5) and the derivative of that is sin(2x^2)*4x -cos(x-5)
Monday, October 18, 2010 at 8:26am by bobpursley
the time index t runs from a to A. if an investment produced a continuous stream of income over 10 years at a rate of $20,000 per year and the interest rate is 6% per year continuously compounded. what is the present value? what is the integral function?
Tuesday, October 19, 2010 at 10:46am by May