Definite integral problem

[cosmic]

Hi guys,

This question is driving me crazy!

I need to find the definite integral from pi/4 to pi/3 of x cos (4x) by using integration by parts. Hence I need the area in 4 d.p. I've managed to get the answer 3/32-(pi sqrt(3))/24 which is approximately -0.132975. Can someone please point me in the right direction. I've spent all day checking it just don't know what I'm doing wrong.

Thanks in advance.

 

[Deleted member 4993]

Hi guys,

This question is driving me crazy!

I need to find the definite integral from pi/4 to pi/3 of x cos (4x) by using integration by parts. Hence I need the area in 4 d.p. I've managed to get the answer 3/32-(pi sqrt(3))/24 which is approximately -0.132975. Can someone please point me in the right direction. I've spent all day checking it just don't know what I'm doing wrong.

Thanks in advance.

What did you get as anti-derivative of:

\(\displaystyle \displaystyle{x * cos(4x)}\)

 

[cosmic]

What did you get as anti-derivative of:

\(\displaystyle \displaystyle{x * cos(4x)}\)

I used the following formula:



The values I got were f(x)=x, g(x)=cos (4x), f'(x)=1 and G(x)=1/4sin(4x).

I think I know where I went wrong. I get to the final stage where I have:


But I know that's wrong. Can you see where I went wrong?

 

[Deleted member 4993]

I used the following formula:

View attachment 4217

The values I got were f(x)=x, g(x)=cos (4x), f'(x)=1 and G(x)=1/4sin(4x).

I think I know where I went wrong. I get to the final stage where I have:

View attachment 4218
But I know that's wrong. Can you see where I went wrong?

View attachment 4217
The formula above is incorrect. Correct formula is

\(\displaystyle \displaystyle{\int u(x) * d[v(x)] \ = \ u(x) * v(x) - \int v(x) * d[u(x)]} \)

In your case:

u(x) = x → d[u(x)] = dx

d[v(x] = cos(4x) dx → v(x) = \(\displaystyle \frac{sin(4x)}{4}\)

 

[cosmic]

View attachment 4217
The formula above is incorrect. Correct formula is

\(\displaystyle \displaystyle{\int u(x) * d[v(x)] \ = \ u(x) * v(x) - \int v(x) * d[u(x)]} \)

In your case:

u(x) = x → d[u(x)] = dx

d[v(x] = cos(4x) dx → v(x) = \(\displaystyle \frac{sin(4x)}{4}\)

I'm sure the formula's correct. That's the notation used in my textbook.

 

[Deleted member 4993]

I'm sure the formula's correct. That's the notation used in my textbook.

You are surely incorrect!!

Follow the instructions I had posted and you'll arrive at the correct answer.

 

[JohnZ]

cos(pi)=-1

 

[HallsofIvy]

I'm sure the formula's correct. That's the notation used in my textbook.

You are surely incorrect!!

Follow the instructions I had posted and you'll arrive at the correct answer.

The two formulas
\(\displaystyle \int_a^b f(x)g(x)dx= \left[f(x)G(x)\right]_a^b- \int_a^b f'(x)G(x)dx\)
(where, I presume, G(x) is an anti-derivative of g()- G'(x)= g(x))
and
\(\displaystyle \int u(x)d(v(x))= uv- \int v(x)d(u(x))\)

are exactly the same with u= f(x) and \(\displaystyle d(v(x))= g(x)dx\)

As JohnZ pointed out, your error is that \(\displaystyle cos(\pi)= -1\) so that \(\displaystyle cos(\frac{4\pi}{3})- cos(\pi)= -\frac{1}{2}- (-1)= \frac{1}{2}\) not \(\displaystyle -\frac{3}{2}\).

 

[cosmic]

Thanks guys.

As John said I had missed out a negative sign.:lol: But the notation I used was correct.