3U Maths Question Thread

[Opengangs]

Hey- I have a few question for integration by substitution.
Could someone please help with these?
   
∫(-2, -3) x(3+x)^7    dx, x=u-3
∫(1,0) x/√(1+x)^3    dx, u+1+x

Btw....how do you use integrals with latex?

sorry its confusing....∫(-2, -3)= the integral between -2 and -3, same for ∫(1,0)= integral between 1 and 0....
thanks for ur help!!

With substitution, there are a few things you need to account for.

1) Bounds change. Because you're going from the \( x \) world to the \( u \) world, the bounds themselves have to be with respect to \( u \), not \( x \).
So when making the substitution, be careful with the bounds.

2) \( du \), not \( dx \). Another common mistake is that they do not account for the changing variable. Because we're making a substitution here, our new integral has to be with respect to the new variable and not \( x \). In order to convert it, we differentiate both sides with respect to the new variable; in this case, it is \( u \). From there, you'll want to rearrange for \( dx \) or some product if it sees fit.

3) Rearranging for \( x \). If (2) does not look feasible, you'll most likely want to rearrange your substitution for \( x \) in order to return a function with respect to \( u \) ONLY. For example, your first question will want you to rearrange for \( x \) to account for the x at the beginning.

So, by considering the substitution we made, we see that:



From there, we just substitute each component in terms of \( u \).

[Never.Give.Up]

With substitution, there are a few things you need to account for.

1) Bounds change. Because you're going from the \( x \) world to the \( u \) world, the bounds themselves have to be with respect to \( u \), not \( x \).
So when making the substitution, be careful with the bounds.

2) \( du \), not \( dx \). Another common mistake is that they do not account for the changing variable. Because we're making a substitution here, our new integral has to be with respect to the new variable and not \( x \). In order to convert it, we differentiate both sides with respect to the new variable; in this case, it is \( u \). From there, you'll want to rearrange for \( dx \) or some product if it sees fit.

3) Rearranging for \( x \). If (2) does not look feasible, you'll most likely want to rearrange your substitution for \( x \) in order to return a function with respect to \( u \) ONLY. For example, your first question will want you to rearrange for \( x \) to account for the x at the beginning.

So, by considering the substitution we made, we see that:



From there, we just substitute each component in terms of \( u \).

Ok thanks for that
Would you be able to explain something like this to me please?

evaluate: the integral bounded by 3 and 0 t/(t+1)^1/2    dt

not sure if this makes sense....

[Opengangs]

Ok thanks for that
Would you be able to explain something like this to me please?

evaluate: the integral bounded by 3 and 0 t/(t+1)^1/2    dt

not sure if this makes sense....

Yep! Sure thing.

So, we need to first choose something to become our substitution. The more you do, the more confident you'll become with it.

\[ \text{Let: } u = t + 1 \]

Now, going through our checklist, we start with the bounds. So, since we're going from the \( t \) world to the \( u \) world, our new bounds have to be with respect to \( u \).
So, we begin with the lower bound: when \( t = 0, u = 0 + 1 = 1 \)
Upper bound becomes: \( t = 3, u = 3 + 1 = 4 \)

Step 2) Now that we've got our bounds out of the way, let's consider how \( t \) differs to \( u \).
Using our substitution, we see that:
\[ u = t + 1 \]
\[ \frac{dt}{du} = 1 \Rightarrow dt = du \]

Step 3) Lastly, we rearrange for \( t \). From our substitution, we see that:
\[ t = u - 1 \]

So, every element of our integral can be rewritten in terms of \( u \)




From here, it's just simple integration.

[Never.Give.Up]

Yep! Sure thing.

So, we need to first choose something to become our substitution. The more you do, the more confident you'll become with it.

\[ \text{Let: } u = t + 1 \]

Now, going through our checklist, we start with the bounds. So, since we're going from the \( t \) world to the \( u \) world, our new bounds have to be with respect to \( u \).
So, we begin with the lower bound: when \( t = 0, u = 0 + 1 = 1 \)
Upper bound becomes: \( t = 3, u = 3 + 1 = 4 \)

Step 2) Now that we've got our bounds out of the way, let's consider how \( t \) differs to \( u \).
Using our substitution, we see that:
\[ u = t + 1 \]
\[ \frac{dt}{du} = 1 \Rightarrow dt = du \]

Step 3) Lastly, we rearrange for \( t \). From our substitution, we see that:
\[ t = u - 1 \]

So, every element of our integral can be rewritten in terms of \( u \)




From here, it's just simple integration.

thanks heaps!

[icandoit]

Hi (:
I am struggling with this parametric question,
P (2ap, ap^2) and Q(2aq, aq^2) lie on the parabola x^2=4ay where a>0. The chord PQ passes through the focus of the parabola.

i) Find a relationship between p and q.
For this, I played around with the gradients of the tangents at P and Q and got: p+q= 2m/a. I am not sure if this is how I am supposed to do it because in the next question it asks,

ii) the tangents at P and Q meet at T. Show that T lies on the directrix.
And when you solve here, you have m as a fourth pronumeral which is annoying. How can I improve my method?

[RuiAce]

Hi (:
I am struggling with this parametric question,
P (2ap, ap^2) and Q(2aq, aq^2) lie on the parabola x^2=4ay where a>0. The chord PQ passes through the focus of the parabola.

i) Find a relationship between p and q.
For this, I played around with the gradients of the tangents at P and Q and got: p+q= 2m/a. I am not sure if this is how I am supposed to do it because in the next question it asks,

ii) the tangents at P and Q meet at T. Show that T lies on the directrix.
And when you solve here, you have m as a fourth pronumeral which is annoying. How can I improve my method?

I'm not sure what your \(m\) is meant to be for part i). If it was meant to be the gradient then it should have disappeared as you progressed through the question.

You should now be able to do part ii).

[Never.Give.Up]

Hello, this further graphs question is bamboozling me...think my derivatives are going wrong somewhere.... i would appreciate any help!!

y= 2x+3/ (x^2-4)

thanks heaps!

[RuiAce]

Hello, this further graphs question is bamboozling me...think my derivatives are going wrong somewhere.... i would appreciate any help!!

y= 2x+3/ (x^2-4)

thanks heaps!

Is this supposed to be \( y = 2x + \frac{3}{x^2-4} \) ?

[Never.Give.Up]

Is this supposed to be \( y = 2x + \frac{3}{x^2-4} \) ?

nope..textbook says



??

[RuiAce]

nope..textbook says



??

Make sure to include all the brackets next time. Without brackets, 2x+3 is interpreted to have the 2x split off from the 3.


Also, you changed from a - to a + in the denominator. Not sure if that was intended but I just used the latest given information

[Never.Give.Up]

Make sure to include all the brackets next time. Without brackets, 2x+3 is interpreted to have the 2x split off from the 3.


Also, you changed from a - to a + in the denominator. Not sure if that was intended but I just used the latest given information

thanks for that....yep sorry will do next time (it was meant to be (x^2-4) my bad...)
how would u go finding 2nd derivative? thats what i seem to be getting confused with..

[Sine]

thanks for that....yep sorry will do next time (it was meant to be (x^2-4) my bad...)
how would u go finding 2nd derivative? thats what i seem to be getting confused with..

To find the 2nd derivitive. First differentiate the original function - this will give you the derivitive.
Now differentiate the derivitive that you've just found - this gives you the 2nd derivitive.

[RuiAce]

Note, however, in general with these types of questions you need not find the points of inflexion. You also should not be using the second derivative to test the nature of a stationary point - it is far easier to just check a bit to both the left and right of the first derivative.

[Never.Give.Up]

Note, however, in general with these types of questions you need not find the points of inflexion. You also should not be using the second derivative to test the nature of a stationary point - it is far easier to just check a bit to both the left and right of the first derivative.

hey thanks for that Rui...we have to find points of inflexion as we are sketching the curve..but, yes i do check nature of stationary point by checking around it...
im just confused as to where the -2 out the front comes from in both first and second derivative..? i must have missed something somewhere....or is it because in the first derivative u have divided it all by negative 2..and just brought it out the front so it just has to carry through into 2nd derivative...?

thanks heaps for ur help and time

[RuiAce]

hey thanks for that Rui...we have to find points of inflexion as we are sketching the curve..but, yes i do check nature of stationary point by checking around it...
im just confused as to where the -2 out the front comes from in both first and second derivative..? i must have missed something somewhere....or is it because in the first derivative u have divided it all by negative 2..and just brought it out the front so it just has to carry through into 2nd derivative...?

thanks heaps for ur help and time

The -2 comes from factorising the numerator. Simplifying \(2(x^2-4)-2x(2x+3)\) gives \(-2x^2-6x-8\), so I just pulled out a -2 in front. Just be careful to double check what’s at the start

For 2U curve sketching, inflexion points are necessary. In general, in fact 90% of the time, for these 3U curves they will not ask for the stationary point. This is because they already know that these will typically be very hard to find. The questions will typically ask for any intercepts, asymptotes and stationary points, but not inflexion points.

In the event where inflexion points are necessary, they will explicitly say so. This will usually be when the inflexion points are easy to find. Just being told to “sketch a curve” in 3U and 4U rarely involves inflexion points