Another question (Integration) (2 Viewers)

[BlueGas]

I need help with question 5. d), I also want to know if the graph I drew was correct for the question or not, thanks.

 

[photastic]

I need help with question 5. d), I also want to know if the graph I drew was correct for the question or not, thanks.

Incorrect graph, there it should pass the origin and looks like a stat pt at x=1 and x= -1. With this question you need to take note of the area under the positive axis.

 

[BlueGas]

Incorrect graph, there it should pass the origin and a jump that looks like a stat pt at x=1. With this question you need to take note of the area under the positive axis.

But I did like a table of values thing where I subbed in x=2, x=1, x=0, x=-1 and x=-2 and the result was the graph I drew. I just looked at the graph on Google now but the results I got from the table of values were plotted on the graph. I just don't get how the graph is wrong even though it is...

 

[enigma_1]

It should look like this: https://www.google.com.au/#q=y=x^3+-x

 

[photastic]

But I did like a table of values thing where I subbed in x=2, x=1, x=0, x=-1 and x=-2 and the result was the graph I drew. I just looked at the graph on Google now but the results I got from the table of values were plotted on the graph. I just don't get how the graph is wrong even though it is...

Always check with wolfram alpha. Here you go http://www.wolframalpha.com/input/?i=y=x((x^2)-1)

 

[enigma_1]

To find the area you basically have to let the area under the x axis be positive aswell and then add all the areas together to get your answer.

 

[BlueGas]

Always check with wolfram alpha. Here you go http://www.wolframalpha.com/input/?i=y=x((x^2)-1)

But in the exam I can't use wolfram how am I meant to know how to graph this sort of question if even the table of values gave me the wrong graph.

 

[enigma_1]

But in the exam I can't use wolfram how am I meant to know how to graph this sort of question if even the table of values gave me the wrong graph.

factorise the equation firstly: y = x(x^2 -1) = x(x-1)(x+1)
from there, you find the x intercepts by letting y=0, so x ints are x= -1, 0 and 1.

Since there is no negative in front of the "x" you know that the graph will be going diagonal from the top right to the bottom left (like a line with positive gradient). Then you basically draw in the middle bit with the line passing through each of the x intercepts with the loops etc.

 

[BlueGas]

But what I don't get is that why couldn't the table of values help me in graphing this sort of question?

 

[photastic]

But what I don't get is that why couldn't the table of values help me in graphing this sort of question?

when x = -2, y = -6
when x = -1, y = 0
when x = -1/2, y = 3/8
when x = 0, y = 0
when x = 1/2, y= -3/8
when x = 1, y = 0
when x = -2, y = 0

Judging by your graph, you missed out what happens when x = - 1/2 and x = 1/2

 

[BlueGas]

when x = -2, y = -6
when x = -1, y = 0
when x = -1/2, y = 3/8
when x = 0, y = 0
when x = 1/2, y= -3/8
when x = 1, y = 0
when x = -2, y = 0

Judging by your graph, you missed out what happens when x = - 1/2 and x = 1/2

I just got told that if a graph has x^3 it starts above the X axis from the right but Google and wolfram say otherwise. I'm kind of confused. ..

 

[photastic]

I just got told that if a graph has x^3 it starts above the X axis from the right but Google and wolfram say otherwise. I'm kind of confused. ..

Why not graph it by finding the stat pts and poi.

 

[InteGrand]

The function is (using difference of two squares). So the polynomial has roots at . also, the function is odd (), so is symmetric about the origin. As . This is enough information to sketch the curve (you don't need to and shouldn't use calculus, since you just need the rough shape).

And there's no stationary points at . all the roots are single roots, so the curve cuts the x-axis at these points like a straight line.

 

[BlueGas]

The function is (using difference of two squares). So the polynomial has roots at . also, the function is odd (), so is symmetric about the origin. As . This is enough information to sketch the curve (you don't need to and shouldn't use calculus, since you just need the rough shape).

And there's no stationary points at . all the roots are single roots, so the curve cuts the x-axis at these points like a straight line.

I understand the polynomial has roots at x=, x=1, x=-1 but what I don't get is how would you know that there are loops on the graph?

 

[enigma_1]

I understand the polynomial has roots at x=, x=1, x=-1 but what I don't get is how would you know that there are loops on the graph?

Since the roots are only -1, 0 and 1 you know that when you're drawing the graph, it's going to cut each point like legit intersect it and go right through it like a loop.
If you have like -1, -1, 0 and 1 as your roots, since there are two -1s, the graph does not cut the x intercept at x= -1 and instead it just touches that point like how the y=x^2 parabola touches the origin.

These are pretty basic concepts which you could look through in your textbook tbh or youtube videos as you need to know how to graph all sorts of functions like second nature.

 

[BlueGas]

Since the roots are only -1, 0 and 1 you know that when you're drawing the graph, it's going to cut each point like legit intersect it and go right through it like a loop.
If you have like -1, -1, 0 and 1 as your roots, since there are two -1s, the graph does not cut the x intercept at x= -1 and instead it just touches that point like how the y=x^2 parabola touches the origin.

These are pretty basic concepts which you could look through in your textbook tbh or youtube videos as you need to know how to graph all sorts of functions like second nature.

I do know how to graph equations when they ask for stationary points, maximum/minimum, pt of inflex, etc. But this type of graphing for integration confuses me, I don't want to be doing the same method of graphing in calculus for integration, it takes up alot of time and I'm sure there's a faster way to graph when answering integration questions.

 

[Carrotsticks]

A problem like question 5 (d) would not be asked in a 2U examination paper anyway because it's a cubic polynomial, the graph of which isn't in the 2U course without the assistance of calculus to sketch it.

 

[enigma_1]

I do know how to graph equations when they ask for stationary points, maximum/minimum, pt of inflex, etc. But this type of graphing for integration confuses me, I don't want to be doing the same method of graphing in calculus for integration, it takes up alot of time and I'm sure there's a faster way to graph when answering integration questions.

For integration it's beneficial to roughly sketch out what the graph would look like. You don't need to figure out stationary points or anything, just find the x intercepts and whether the leading coefficient is positive or negative and whether there is a y intercept. Using all that info and how many roots the equation has, you can draw up a quick sketch of any graph for integration purposes to visualise what's happening.

 

[BlueGas]

A problem like question 5 (d) would not be asked in a 2U examination paper anyway because it's a cubic polynomial, the graph of which isn't in the 2U course without the assistance of calculus to sketch it.

So basically any equation that has a cubic polynomial won't be in the 2U exam? That's a relief. Because I know how to sketch x^2 and anything above I can't.

 

[BlueGas]

Also questions like 3. b), 3. f) and 5. c) won't be in the 2U exam because there cubic polynomials?