Your personal IBP Problem

(10 pts)

Enter your forum login name into the text box below and hit the “Submit” button. A WebWork problem on Integration by Parts should automagically appear. Your mission is to do the problem and explain how you did it in a forum response.

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• A function like x\sin(x) should be entered as $x\sin(x)$, and
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$$
\int_0^{2\pi} \sin(x) dx = 0.
$$

I’m asked to compute

To do so, I’ll choose f(z) = z+1 and g'(z) = e^{8z}.
Then we have f'(z) = 1 and g(z) = e^{8z}/8, so

My integral was;

\int x\cos(46x)dx.

To start, have f(x)=x and g'(x)=\cos(46x).
Then, f'(x)=1 and g(x)=\frac{1}{46} \sin(46x).

Finally, apply by parts formula; f(x)g(x)- \int f'(x)g(x) and simplify.

I got the Problem

I used f(x) = x^2 and g'(x) = cos(2x).
Which makes f'(x) = 2x and g(x) = \sin(2x)/2, so

My integral was:

When I rewrote it as

it became clear that my integral was of the form \int\,{x^{p}\ln x}\,dx.

So, I assigned my parts like this:

\mathrm{let\ }f(x) = \ln x
\mathrm{let\ }g'(x) = x^{-4}

Then I found f'(x) and g(x):

f'(x) = x^{-1}
g(x) = - \frac{1}{3}x^{-3}

I plugged my parts into the IBP formula \int\,UdV = UV - \int\,VdU to get:

I simplified the integral:

And then solved it with the reverse power rule:

Then I simplified my answer (because “style counts”):

My Intergal was:

\displaystyle\int xe^{2x} \, dx

I used Intergation by parts with f(x) = x and g'(x)=e^{2x}

with f'(x) = 1 and g(x)=\frac{e^{2x}}{2}

and when plugged into the formula I got \displaystyle\int xe^{2x} \, dx = x\frac{e^{2x}}{2}-\displaystyle\int 1*\frac{e^{2x}}{2} \, dx.

For \frac{e^{2x}}{2} I used u-sub with u= 2x which gave me the integral \frac{1}{4}\displaystyle\int e^{u} \, dx.

The integral of \displaystyle\int e^{u} \, dx is e^u which is e^{2x} after you reverse the subution, so \frac{1}{4}\displaystyle\int e^{u} \, dx = \frac{e^{2x}}{4}.

And thus my answer is x\frac{e^{2x}}{2}-\frac{e^{2x}}{4}+C

My integral was

I first assigned my f(x), f’(x), g’(x) and g(x) values:

Then, I plugged them into the integration by parts formula:

Which resulted in:

After simplifying that, I ended up with:

Evaluated from 0 to 1. I cannot figure out how to symbolize this with the coding.

I then used the fundamental theorem of calculus to solve for the bounds of integration:

The integral that I received was:

after moving the constant the integral became

and I chose f(x) = y and g'(x) = e^{-2y}

then the problem becomes f'(y)=1 and g(y)=-\frac{1}{2}{e^{-2y}}

After applying the parts formula f(x)g(x)- \int f'(x)g(x)

after some simplification the problem becomes

Using the Fundamental Theorem of Calculus and evaluating the problem on the integral \int_0^{1}

The final answer equates to

The problem I received was:

The first step is to assign f(x) and g'(x). I opted to assign ln(2t) to f(x) and t^4 to g'(x). Then calculate f'(x) and g(x) which are (1/t) and (1/5)t^5 respectively.

Next combine the parts as stated in IBP to get:

Clean up the integral by canceling a t:

Integrate and you’re left with:

Combine and clean up to get:

Now add in the upper and lower bounds to find the final answer:

If you’re curious, in decimal form this comes out to about 8198.5487 rounded.

I was given

I decided
f(x)=x therfore f'(x)=1 and
g'(x) = e^{{3x}} therfore g(x)= e^{{3x}}/3 so,

My problem was:

\int xcos(24x)dx

I made f(x)=x and g'(x)=cos(24x)
This allowed f'(x)=1 and g(x)=\frac{1}{24}sin(24x)

This would allow the equation to be written as
x(\frac{1}{24}sin(24x))-\frac{1}{24}\int sin(24x)dx

Then using u-substitution:
let u=24x and du=24dx making \frac{1}{24}du=dx

This makes the second half of the equation look like this:
\frac{1}{24}\int sin(u)\frac{1}{24}du

Then after rearranging the equation it should look like this:
\frac{1}{24}(\frac{1}{24}\int sin(u)du)

To simplify:
\frac{1}{576}\int sin(u)du

Then take then integral to get:
\frac{1}{576}-cos(u)

Then plug in u to get:
\frac{1}{576}-cos(24x)

To get the final answer, plug the answer from the u-sub problem into the integral by parts problem:
\frac{1}{24}xsin(24x)+\frac{1}{576}cos(24x)+C

I was given:

To do this, I chose:
\ u= z + 1 and \ dv= e^{5z}
Therefore, \ du=dz, and \ v = (1/5)e^{5z}

From there, I used the formula \int\ udv = uv - \int\ vdu

So I plugged in those values to get:

then solved the simpler integral to find the solution:

My integral was:

I choose f(t)=ln(t) and g'(t)=\sqrt(t). So we also have f'(t)=\frac1t and g(t)=\frac23 \sqrt(t^3).
Now we use our formula f(t)g(t)-\int f'(t)g(t) and so we have
2(\frac23 ln(t)\sqrt(t^3) - \int \frac23 \sqrt(t) dt) which can be simplified into
(\frac43ln(t)\sqrt(t^3) - \frac89 \sqrt(t^3)