Mathematics Question Thread

[fun_jirachi]

And you can see only the first four solutions fit the domain, so that's where the answer comes from. I'm kind of curious as to how you got your answer, since it looks as if your answers are solving for sin 2x = 0, not 3cos(2x +π/4) = 0. Maybe share your working out, and we can see where you went wrong specifically!

[emilyyyyyyy]

Hi,

Can someone please help me with the two questions attached?

thanksssss

[benneale]

hey! please help...

evaluate:
bounds 2 and 1 for 2x^3-x^2+5x+3 all over x, dx

[RuiAce]

Hi,

Can someone please help me with the two questions attached?

thanksssss

\[ \text{Since you have the sketch of the cubic given you know your integration boundaries.}\\ \text{To do the integral, you can just expand the brackets to get}\\ y=x (-x^2+5x-6) \implies \boxed{y = -x^3+5x^2-6x} \]
\begin{align*}\therefore A&= \left| \int_0^2 -x^3+5x^2-6x\, dx \right| + \int_2^3 -x^3+5x^2-6x\,dx \end{align*}
\[ \text{The rest is left as your exercise to compute.} \]
With regards to the second question, it technically is not a part of the 2U course. However some textbooks teach the rule \( \int f^\prime(x) e^{f(x)}dx = e^{f(x)}+C\) anyway. If your textbook does this, you should consider the fact that \( \int (x^2+1)e^{x^3+3x}dx = \frac13 \int (3x^2+3)e^{x^3+3x}dx \), and try applying the formula from there.

hey! please help...

evaluate:
bounds 2 and 1 for 2x^3-x^2+5x+3 all over x, dx

I don't see where the main difficulty is in this question. Please post relevant working out or add specifically where the problem is.

[emilyyyyyyy]

\[ \text{Since you have the sketch of the cubic given you know your integration boundaries.}\\ \text{To do the integral, you can just expand the brackets to get}\\ y=x (-x^2+5x-6) \implies \boxed{y = -x^3+5x^2-6x} \]
\begin{align*}\therefore A&= \left| \int_0^2 -x^3+5x^2-6x\, dx \right| + \int_2^3 -x^3+5x^2-6x\,dx \end{align*}
\[ \text{The rest is left as your exercise to compute.} \]
With regards to the second question, it technically is not a part of the 2U course. However some textbooks teach the rule \( \int f^\prime(x) e^{f(x)}dx = e^{f(x)}+C\) anyway. If your textbook does this, you should consider the fact that \( \int (x^2+1)e^{x^3+3x}dx = \frac13 \int (3x^2+3)e^{x^3+3x}dx \), and try applying the formula from there.

thank you!
I've attached my working for the area question and i cant seem to get the right answer! What have I done wrong in my working out?

[RuiAce]

thank you!
I've attached my working for the area question and i cant seem to get the right answer! What have I done wrong in my working out?

Not too sure why you only put on absolute values at the end, instead of around only the first integral like what I wrote down?

Recall that absolute values are only used for regions below the \(x\)-axis in order to focus on the magnitude of the area only. By essentially ignoring the absolute values, you end up calculating the sum of the signed areas instead, i.e. in the process of doing so you treat the area under the \(x\)-axis as negative.

[emilyyyyyyy]

Hi I need help again!

I'm having trouble solving questions like the one attached with x's in the numerator!
How do I solve it?

thanks

[emilyyyyyyy]

Not too sure why you only put on absolute values at the end, instead of around only the first integral like what I wrote down?

Recall that absolute values are only used for regions below the \(x\)-axis in order to focus on the magnitude of the area only. By essentially ignoring the absolute values, you end up calculating the sum of the signed areas instead, i.e. in the process of doing so you treat the area under the \(x\)-axis as negative.

my bad! i put them around the first one instead and my answer is 37/12, but the I'm still not getting the answer.

I've attached the solution given to me, but I don't really understand how they did it.

[fun_jirachi]

Hi I need help again!

I'm having trouble solving questions like the one attached with x's in the numerator!
How do I solve it?

thanks

In general, you want to be splitting up the integral and manipulating it into a better form to integrate.

[RuiAce]

my bad! i put them around the first one instead and my answer is 37/12, but the I'm still not getting the answer.

I've attached the solution given to me, but I don't really understand how they did it.

That looks like the solution to a completely different question to me.

I checked on Wolfram and the answer to the one you posted is definitely \( \frac{37}{12} \).

[emilyyyyyyy]

Hi,

How do I integrate logs without fractions?
For e.g. y=ln(x-2) ?

thanks!

[slinkybench]

I can't figure out how to do this question: Consider the two simultaneous equations mx + y = 24 & 6x + my = m. Find the exact values of m for which there is no solution to the pair of simultaneous equations.

Thanks in advance.

[RuiAce]

Hi,

How do I integrate logs without fractions?
For e.g. y=ln(x-2) ?

thanks!

Short answer: You don't (in 2U).

Long answer: You still don't, but if you're given a definite integral in 2U you can potentially draw a picture and consider something involving \( \int_a^b e^y\,dy\). Depends on the question at hand.

I can't figure out how to do this question: Consider the two simultaneous equations mx + y = 24 & 6x + my = m. Find the exact values of m for which there is no solution to the pair of simultaneous equations.

Thanks in advance.

\[ \text{For a purely algebraic approach, we can attempt to start solving.}\\ \text{The first equation rearranges to }y=-mx+24.\text{ So subbing into the second gives} \\ \begin{align*} 6x+m(24-mx) &= m\\ 6x + 24m - m^2x &= m\\ x(6-m^2) &= -23m\\ x &= \frac{-23m}{6-m^2}.\end{align*} \]
\[ \text{From here, we see that we will not have any solutions if }6-m^2=0\\ \text{which solves to give }\boxed{m=\pm \sqrt{6}}. \]
Note that this essentially boiled down to when do we have "divide by zero" problems. i.e. when the denominator is zero. Because then the result is undefined.

[emilyyyyyyy]

Hi,

With this question, when I'm working it out, what am I meant to do to know whether the square root of y is + or - in the graph?

Thanks!

[benneale]

hey, having a bit of a 'moment'...

Find the arc length, correct to 2 decimal places, given radius is 5.9cm and angle subtended is 23degrees 12minutes...

I understand the process of l=r.theta , but when I punch in the degrees bit into my calculator, the answer is way off!