Algebra &amp; Calculus questions

Original post by SubAtomic

Yeah I get that, it is a many one function though and what I get after all the calculating is incorrect. I already know the solution to the integral as have differentiated a similar equation already, something is not working out when I do the

\displaystyle [F(x)]^b_a

May be some of the area is below the x-axis, so you are getting the 'net area'. You need to change the limits in this case.

Reply 62

If I was to have an integral like this

\displaystyle\int\dfrac{18 \cdot (x^n+2)}{7(x-1)} \ dx

Then I could take out the 18 and leave the 7 in?

\displaystyle 18 \cdot \int\dfrac{(x^n+2)}{7(x-1)} \ dx

Or if the integral was

\displaystyle\int\dfrac{18 \cdot (x^n+2)}{(x-1)} \ dx

but I needed a 7 in the denominator then I would have

\displaystyle\int\dfrac{126 \cdot (x^n+2)}{7(x-1)} \ dx

and could take the 126 outside the integral

\displaystyle 126 \cdot \int\dfrac{(x^n+2)}{7(x-1)} \ dx

(edited 11 years ago)

Reply 63

Original post by SubAtomic

If I was to have an integral like this

\displaystyle\int\dfrac{18 \cdot (x^n+2)}{7(x-1)} \ dx

Then I could take out the 18 and leave the 7 in?

\displaystyle 18 \cdot \int\dfrac{(x^n+2)}{7(x-1)} \ dx

Or if the integral was

\displaystyle\int\dfrac{18 \cdot (x^n+2)}{(x-1)} \ dx

but I needed a 7 in the denominator then I would have

\displaystyle\int\dfrac{126 \cdot (x^n+2)}{7(x-1)} \ dx

and could take the 126 outside the integral

\displaystyle 126 \cdot \int\dfrac{(x^n+2)}{7(x-1)} \ dx

The answer to both of your question is yes.

Reply 64

Original post by SubAtomic

If I was to have an integral like this

\displaystyle\int\dfrac{18 \cdot (x^n+2)}{7(x-1)} \ dx

Then I could take out the 18 and leave the 7 in?

\displaystyle 18 \cdot \int\dfrac{(x^n+2)}{7(x-1)} \ dx

Or if the integral was

\displaystyle\int\dfrac{18 \cdot (x^n+2)}{(x-1)} \ dx

but I needed a 7 in the denominator then I would have

\displaystyle\int\dfrac{126 \cdot (x^n+2)}{7(x-1)} \ dx

Yes, you can do this.

And you can also take both out.

\displaystyle\int\dfrac{18 \cdot (x^n+2)}{7(x-1)} \ dx = \displaystyle 18 \cdot \int\dfrac{(x^n+2)}{7(x-1)} \ dx = \displaystyle \frac17 \int\dfrac{18(x^n+2)}{(x-1)} \ dx = \frac{18}7 \int\dfrac{(x^n+2)}{(x-1)} \ dx

Reply 65

Original post by SubAtomic

Cheers, no idea what is going on with my definite integral then :s-smilie:

You need to give us the question, so that we can see what is going on with it.

Otherwise, like before, wait for the dream.

Reply 66

Original post by raheem94

You need to give us the question, so that we can see what is going on with it.

Otherwise, like before, wait for the dream.

You aren't very creative are ya :tongue:

Reply 67

Original post by SubAtomic

PS Raheem what is this talk of changing the bounds? I need to work out above x-axis so below is irrelevant

You were having some problems finding the area by integration.

In integration, area below the x-axis in -ve, so i was suggesting that the graph of your function, might also be in the -ve region.

Reply 68

Original post by SubAtomic

You aren't very creative are ya :tongue:

So true

Reply 69

So let me get this air tight as am going to have to perservere with this, little different example

\displaystyle\int\dfrac{18 \cdot (x^n+2)y}{(x-1)} \ dx

but I needed a 7 in the denominator then I would have

\displaystyle\int\dfrac{126 \cdot (x^n+2)y}{7(x-1)} \ dx

and could take the 126 outside the integral

\displaystyle 126 \cdot \int\dfrac{(x^n+2)y}{7(x-1)} \ dx

Plus in general when I ask a question I am a bit cryptic about you can look at it as me not posting the answers to an exam you are sitting tomorrow type situation. Sorry if this is annoying but it is the way it has to be.

(edited 11 years ago)

Reply 70

The confusion has come from me using integers instead of radians for the stationary points :rolleyes:

Reply 71

Original post by SubAtomic

So let me get this air tight as am going to have to perservere with this, little different example

\displaystyle\int\dfrac{18 \cdot (x^n+2)y}{(x-1)} \ dx

but I needed a 7 in the denominator then I would have

\displaystyle\int\dfrac{126 \cdot (x^n+2)y}{7(x-1)} \ dx

and could take the 126 outside the integral

\displaystyle 126 \cdot \int\dfrac{(x^n+2)y}{7(x-1)} \ dx

That's also correct.

Reply 72

Original post by notnek

That's also correct.

So the area under a graph a continuous function

\displaystyle f(x) =[F(x)]^b_a

Something isn't working out, have looked at some definite integrals in an under grad book, looked through my course books and am still non the wiser.

Will have a go at one here but do not have a clue why this isn't applicable to the one I am stuck on, maybe I need to have four different intervals within the domain, like [a,c] and [c,b] :s-smilie:

\displaystyle \int_0^{\pi} (\cos (3x)+6 \sin (4x)) \ dx

\displaystyle = \left[\frac{1}{3} \sin(3x)-\frac{3}{2} \cos(4x) \right]_0^{\pi}

\displaystyle = \left(\frac{1}{3} \sin(3 \pi)- \frac{3}{2} \cos(4 \pi)\right) - \left(\frac{1}{3} \sin(0) - \frac{3}{2} \cos(0)\right)

Need to pop out now but will come back and edit this in a bit, please point out any mistakes in the meantime.

\displaystyle = \left(0- \frac{3}{2} \right)- \left(0-\frac{3}{2} \right) = 0

(edited 11 years ago)

Reply 73

[QUOTE="SubAtomic;37436262"]So the area under a graph a continuous function

\displaystyle f(x) =[F(x)]^b_a

\displaystyle \int_0^{\pi} (\cos (3x)+6 \sin (4x)) \ dx

\displaystyle = \left[\frac{1}{3} \sin(3x)-\frac{3}{2} \cos(4x) \right]_0^{\pi}

\displaystyle = \left(\frac{1}{3} \sin(3 \pi)- \frac{3}{2} \cos(4 \pi)\right) - \left(\frac{1}{3} \sin(0) - (\frac{3}{2} \cos(0)\right)

Need to pop out now but will come back and edit this in a bit, please point out any mistakes in the meantime.

\displaystyle =

It's all good so far.

By the way, when you say, "it's not working out", how do you know?

Reply 74

Original post by notnek

It's all good so far.

By the way, when you say, "it's not working out", how do you know?

As you can see now have edited it I am making a newb error somewhere surely, and I have used mathcad to find the increased function so that is where my suspicions come from.

(edited 11 years ago)

Reply 75

Original post by SubAtomic

As you can see now have edited it I am making a newb error somewhere surely, and I have used mathcad to find the increased function so that is where my suspicions come from.

Ah I see your problem. When you find this integral between 0 and pi, the 2 regions above the axis cancel with the two regions below the axis (you can see this by looking at the graph), to leave you with 0. So your calculations were correct.

Which area are you looking to find?

Reply 76

Original post by notnek

Ah I see your problem.
Which area are you looking to find?

either would be good lol, they are both identical other than the negative sign aren't they? What am I doing wrong?

(edited 11 years ago)

Reply 77

Original post by SubAtomic

either would be good lol, they are both identical other than the negative sign aren't they, what am I doing wrong?

You may want to find the total area i.e. the sum of the areas of the four regions.

If you want to find the area above the x-axis, then you need to consider the two regions separately. And for that you'd need to know where the graph crosses the x-axis. Once you know the two regions, [a,b] and [c,d], then the area would be:

\displaystyle \int_a^b \cos(3x)+\sin(4x) + \int_c^d \cos(3x)+\sin(4x)

Is this still an example question? I don't think you'd have this problem with the question you PM'd me.

Reply 78

Original post by notnek

Is this still an example question? I don't think you'd have this problem with the question you PM'd me.

Yep still an example of sorts to make sure am doing things "right", and yep get ya the function I pm'd you doesn't cross the x-axis in the interval given so I have no idea what is going on.

What module of the A-level syllabus would you say the question I pm'd is similar to (taken from)?

(edited 11 years ago)

Reply 79