knowing the standard form for integration by parts is
∫ f(x)g'(x) dx = f(x)g(x) - ∫f'(x)g(x) dx
I have what is an innocuous looking part of an equation which I can't solve.
the f(x) part in this case is;
ln(5x) which is easy enough i.e. 1/x
the second part 1/(x(2/3)) is the bit I...
greetings . it's known that if g(x), f(x) are two functions ,and f(x) is sufficiently differentiable , then by repeated integration by parts one gets :
\int f(x)g(x)dx=f(x)\int g(x)dx -f^{'}(x)\int\int g(x)dx^{2}+f^{''}(x)\int \int \int g(x)dx^{3} - ...
Homework Statement
Find the integral of 3x* (2x-5)^6*dx, let u= 2x -5.
Homework Equations
Im not sure if i am meant to use integration by parts or not?? I was able to do previous questions of the topic just using u sub to get rid of the first x variable.
The Attempt at a Solution...
Homework Statement
Evaluate the integral.
Integral = x f(x) dx from 0 to 1 when f(1) = 6, f'(1) = 7.
Answer choices:
A. 11/6 + 1/6 integral from 0 to 1 x^3f''(x)dx
B. 11/12 - 1/6 integral from 0 to 1 x^3f''(x)dx
C. 11/3 + 1/2 integral from 0 to 1 x^2f'(x)dx
D. 11/3 - 1/2 integral from 0 to 1...
I found this interesting but different way to solve integration by parts problems on the internet: http://imageshack.us/photo/my-images/854/integration20by20parts2.jpg/
It seems to work well for me when doing most textbook problems, except when the integrand contains a natural logarithm. I just...
I'm confused. I was making up some of my own problems involving higher powers of x to integrate. For example:
\displaystyle\int x^5 e^{5x}dx
I set about going about finding \frac{dy}{dx} up to \frac{d^6y}{dx^6}.
u=x^5
\frac{du}{dx}=5x^4
\frac{d^2u}{dx^2}=20x^3
\frac{d^3u}{dx^3}=60x^2...
Homework Statement
Here are two instances where the negative sign just changes for no reason. The one's all the way on the right. Why? I don't understand what is going on here. For the second one, it should + cos x
Homework Statement
∫ x2 sin x
Homework Equations
uv - ∫ v duThe Attempt at a Solution
u = x2
du = 2x
dv = sin x
v = -cos x
step 1. x2 - cos x - ∫ -cos x 2x
I think -cos x * 2x becomes -2x cos x
so now we have
step 2. x2 - cos x - ∫ -2x cos x
which means I have to integrate by parts...
Homework Statement
In this video from which there is a screen shot above the author went from x/2 to 2x and all he said was half is two quarters. right a half is two quarters it is not 2. I just want to make sure that he made a mistake because I've been seeing some real bizarre things in...
Homework Statement
integrate .5e^(t/50)*sin(t)
Homework Equations
integration by parts
uv-∫vduThe Attempt at a Solution
I am currently in differential equations and I remember from cal II that I have to keep using the equation above until the integral loops around, then set it equal to...
Homework Statement
Integrate By Parts (i.e. not using formulas)
∫e3xcos(2x)dx
The Attempt at a Solution
I keep going around in circles, I know at some point I should be able to subtract the original integral across the = and then divide out the coefficient and that's the final...
Hello. I'm attempting to integrat ∫ln(x+x^2)dx
Our professor gave us the hint of x(1+x)
I believe u= ln(x+x^2) and du=1+2x/x+x^2
I am not sure what dv should be
Any help would be greatly appreciated! Thanks
Homework Statement
Hi,
I'm doing fouier transforms and I'm not sure how to integrate (2-x)cos(nPi/2)x, (1,2). Anyone able to help me out? Even the indefinite integral would be fine.
Homework Equations
The Attempt at a Solution
I guess u would be (2-x) and dv would be cos(nPi/2)x dx. I'm not...
Substitution method with Integration by Parts?
Homework Statement
Evaluate the integral...
∫x^3[e^(-x^2)]dx
Homework Equations
∫udv=uv-∫vdu
The Attempt at a Solution
I first tried using integration by parts setting u and dv equal to anything and everything. This seemed to make...
∫xax
u=x
du=dx
dv=axdx
v=ax/lna
= xax - ∫axdx/lna
is my solution right?
my problem now is how to integrate the expression xax - ∫axdx/lna
please help..
Homework Statement
Calculate the following integral exactly (no approximations) by the method of integration by parts:
∫0t SinIntegral[x] dx
Homework Equations
the following hints are given:
D[SinIntegral[x], x] = Sinc[x]; and
SinIntegral[0] = 0
The Attempt at a Solution...
Homework Statement
Use integration by parts to derive the formula:
\int (a^2 - x^2)^n dx = \frac{x(a^2-x^2)^n}{2n+1} + \frac{2a^2n}{2n+1}\int \frac{(a^2 - x^2)^n}{(a^2 - x^2)} dx + C
Homework Equations
Integration by parts general formula
∫udv = uv - ∫vdu
The Attempt at a...
Hi, I'm wondering how to integrate 4xe^(4x).
I got:
4[1/4xe^(4x)-1/16e^(4x)+c] ?
which reduces to
xe^(4x)-1/4e^(4x)+c
Is this the correct integral?
Thanks.
Homework Statement
Solve integral
\int^{1}_0(1-x)\frac{d}{dx}\frac{\sin Cx}{C}dx
Homework Equations
\int udv=uv-\int vdu
The Attempt at a Solution
u=1-x
dv=\frac{d}{dx}\frac{\sin Cx}{C}dx
What is v?
How to integrate
\frac{d}{dx}\frac{\sin Cx}{C}dx?
Homework Statement
I am doing self-study. I am on problem 5.6 #27 in the Stewart text 3rd E.
I don't understand why the integration limits changed after the given substitution.
The given substitution was:
x=θ^2 dx=2θdθ
Homework Equations
Please see attachment.
The Attempt at...
Homework Statement
\int_0^\infty{ \frac{1}{x} e^{-x}}
Homework Equations
Integration by parts
\int{u dv} = uv - \int{v du}
The Attempt at a Solution
u = \frac{1}{x}
du = \frac{1}{x^2} dx
v = -e^{-x}
dv = e^{-x} dx
-\frac{1}{x} e^{-x} - \int_0^\infty{-e^{-x}...
I'm watching the video series on Quantum Mechanics taught by Leonard Susskind, (from Stanford).
On Lecture #3, Dr. Susskind says that integration by parts is:
∫FG' = -∫GF'
However from what I know integral by parts to be, there i missing a +FG on the righthand side, or something... since I...
Hello, thanks for reading.
This is a general question: as far as I know, integration by parts is allowed only with functions that are continuously differential.
However, I'm reading Griffiths Quantum book, and he easily uses this technique in integrals involving the delta "function" and the...
Homework Statement
For ordinary 1D diffusion show that the mean value of the square of the position is equal to 2Dt
Homework Equations
\left\langle {x^2 \left( t \right)} \right\rangle \equiv \int\limits_0^\infty {x^2 p\left( {x,t} \right)dx}
\frac{\partial }{{\partial t}}p\left(...
Homework Statement
∫▒〖(2x-1)e^(-x) 〗 dx
I don't want to butcher this but I know you use integration by parts, I just don't know how to do this one in particular because i is one of the simple ones I was told. Please Help
Homework Statement
problem: \int2arctanx dx
2\intarctan dx
u=arctanx
du=1/(1+x2)
v=x
dv=dx
xarctanx-\intx/(1+x2)
integrate by parts a second time...
u=x
du=dx
v=arctanx
dv=1/1+x2
xarctanx-\intarctanx
My final answer I get it 2xarctanx-2xarctanx+2/x2+1 which is...
What is the difference? I was pretty bored last night so I got onto Yahoo Answers and answered a few calculus questions. It was a simple integration by parts question:
\intxsin(x) dx
I solved as:
u = x
du = dx
dv = sin(x) dx
v = -cos(x)
uv - \intvdu
-xcos(x) + \intcos(x)dx =...
Here I used integration by parts to try to solve an integral (I got it wrong, it seems), I know this has no "simple" solution, but, can anyone explain me exactly what did I do wrong? Here is what I did...
Homework Statement
Hi
There is a step in my book, which I can't follow. It is the following
\int_0^1 {w\left( {\frac{{d^2 u}}{{dx^2 }} - u + x} \right)dx} = \int_0^1 {\left( { - \frac{{dw}}{{dx}}\frac{{du}}{{dx}} - wu + xw} \right)dx} + \left[ {w\frac{{du}}{{dx}}} \right]_0^1
I...
Homework Statement
Find the solution to:
y' = x.y.cos(x^2)Homework Equations
Integration by Parts method.The Attempt at a Solution
Step 1
(dy/dx).(1/y) = x.cos(x2)
(1/y) dy = x.cos(x2) dx
Step 2
Integrate both sides.
ln|y| = integratal of [ x.cos(x2) dx ]
Step 3
Using integration by...
The attempt at a solution
\begin{equation*}
\begin{split}
\ -\ i\int\psi^* \frac{\partial{\psi}}{\partial x}= \\ -i(\psi^*\psi\ - \int\psi \frac{\partial\psi^*}{\partial x})\space\ (?)
\end{split}
\end{equation*}
I thought \psi^*\psi\ = \ constant\neq\ 0. However, it vanishes in...
I'm just curious about the proofs of Integration by Parts & the Change of Variables formula
as given in this book on page 357. I think there are a lot of typo's so I've uploaded my
rewrite of them but I am unsure of how correct my rewrites are. If someone could point
out the errors & why I...
Basically I have answered a question using the integration by parts formulae to work out the centre of gravity inside a fan blade using :-
v.du/dx = v.u - u. dv/dx with the integral limits of 0 ==> 20
when v = x then dv/dx =1
when du/dx = 0.3 sinx then u = 0.3cos x
and sub this into...
Homework Statement
Take the integral of the following:
1. ln(2x+1)
2. arctan4x
3. ecosxsin2x evaluated from 0 to pi
The Attempt at a Solution
1. took the derivative of ln(2x+1) and integrated dx. my solution was:
xln(2x +1) + x + [(2x + 1)-2]/2 + C
The books answer was...
Homework Statement
\int xarcsin2xdx
2. The attempt at a solution
Can someone explain to me what is happening at step 2? I understand how the integration by parts was done, but where does the (1/8) or (2x) come from?
Homework Statement
\int\sqrt{4+9x^{2}}dx
Homework Equations
Pythagorean Identities?
The Attempt at a Solution
I find it sort of cumbersome to use the special formatting here, so I hope it is okay that I just photocopied my work on paper.
You can see how far I made it, but...
1. How to solve integral of (1/(t2-t))dt
2. to be solved without using laplace transforms
3. integral of( uv)= u*(integral of v) -integral of ((u')*(integral of v)) ... right?
integral of (1/t^2-t) = integral of (1/t)*(1/t-1)dt = (1/t-1)*(log t) - integral((-1/(t-1)2*logt ...i don't...
Homework Statement
integrate (x*2e^x)/(2e^x-1)2 from x=0 to infinity
Homework Equations
The Attempt at a Solution
let t=2e^x-1 => x=ln((t+1)/2)
dt = 2e^x dx
Thus equation is now integrate (ln((t+1)/2))/t^2 dt from t=1 to infinity
Then let u = (t+1)/2 => 2du=dt
Equation now...
I am assuming that the solution was arrived at through integration by parts, however I am not able to completely work through it.
First given: cB= XB/Vm
the next step shows the solution to dcB given as:
dcB=(1-dlnVm/dlnxB)(dxB/Vm)
Homework Statement
i)Use integration by parts to express:
I(n) = ∫ sin^n (x) dx
in terms of I(n-2).
ii) Hence show that ∫(π/2 for top, π/4 for bottom) 1/[sin^4 (x)] dx = 4/3
Homework Equations
Reduction Formula and Trig Identity [sin²(x) + cos²(x) = 1]
π = pi
The Attempt at a...
Homework Statement
i have to create a general formula for integral of (x^n * e^x) dx
using whatever method i deem appropriate. (the only way i could think of is by parts)
Homework Equations
int(x^n * e^x)dx
int(uv')dx=uv-int(vu')dx
The Attempt at a Solution
i used integration by...
Homework Statement
∫ x * e^-x dx
Homework Equations
Integration by parts: Just wondering if below is correct. Not brilliant with Integration by parts and not sure if my +ve and -ve signs are correct. Some help to say if i am correct or where i have gone wrong would be brilliant...
[b]1. The problem statement, all variables and given/known
Homework Statement
\int \frac{sinx}{x}dx
Homework Equations
The Attempt at a Solution
Which method should work here? I tried integration by parts and it looks too much.
Is there a way to solve it without approximating it with the...
Homework Statement
I had this integral on my physics homework and for the life of me couldn't solve it. I ended up using Maple..well wolframalpha.com because Maple's output sucks.
Anyway here is the problem.
\int_{0}^{\infty} x e^{-2 \alpha x}dx
Homework Equations
\int u dv = uv - \int v...
I would like to solve the following integral but I am unsure of the best way to solve it:
\int_{0}^{H}xsin(\frac{w}{x})cos(\frac{x}{w})cosh(\frac{H}{w})dx
Is it possible to use integration by parts??
Thanks in advance
Hi All,
This is not a homework question, I am just trying to be come quicker at integrating by parts, when performing Laplace Transforms.
My textbook gives a basic example for performing the Laplace Transform of the variable t, to the transformed variable of s for the
equation...
the expression to integrate is:
\int x^{3}e^{x^{2}}dx
and in the spirit of "LIATE" I set my u and dv as the following:
dv=e^{x^{2}}dx
u=x^{3}
however, doing this that I integrate dv=e^{x^{2}}dx in order to get v...and unless I'm missing something, this does not seem like an easy...
problem is to integrate the following by parts:
\int x\sec^{2}xdx
my feeling is convert the secant term to cosine by:
sec^{2}x=cos^{-2}x\Rightarrow\int\sec^{2}xdx=\int\cos^{-2}xdx
then:
u=\cos^{-2}x\implies du=2\sin x(\cos^{-3}x)
and also:
dv=xdx\implies v=\frac{x^{2}}{2}...