Thread: Some questions about differentiation/ integration

The first should be pretty simple. But I'm just checking to see if I made any careless mistakes. My answer is -x/4+1/2.

The second question seems tricky. I thought about doing it the manual way ie.e differentiating and integrating y to find it's first derivative, then divide it by 2, where I'll obtain g(x) and I' d then integrate g(x) with limits 2 and 1 respectively. Is there an easier way? Or is my idea above flawed? I don't have the answer with me at the moment.

The third should be similar. But if I integrate the first derivative to obtain it's original equation, I'd have to add a constant value to the equation. Unless I don't need to do so, I should be able to obtain y=14x, where I integrate it with limits 3 and 0. Is that correct?

Regarding the fourth question, I'm not sure on how to proceed. Does the first part of the question refer to 0.5(3+8)(6)?

Originally Posted by smmxwell

The first should be pretty simple. But I'm just checking to see if I made any careless mistakes. My answer is -x/4+1/2.

The second question seems tricky. I thought about doing it the manual way ie.e differentiating and integrating y to find it's first derivative, then divide it by 2, where I'll obtain g(x) and I' d then integrate g(x) with limits 2 and 1 respectively. Is there an easier way? Or is my idea above flawed? I don't have the answer with me at the moment.

The third should be similar. But if I integrate the first derivative to obtain it's original equation, I'd have to add a constant value to the equation. Unless I don't need to do so, I should be able to obtain y=14x, where I integrate it with limits 3 and 0. Is that correct?

Regarding the fourth question, I'm not sure on how to proceed. Does the first part of the question refer to 0.5(3+8)(6)?

1.- dy/dx = (dy/dt)(dt/dx) = (1 - 4t)(1/4) = 1/4 - t = -x/4 + 1/2 and your solution's correct.

2.- Of course there's a much simpler way to do this: g(x) = 1/2(dy/dx) = (1/2)y' dx, so:

INT{1,2}[g(x)dx] = (1/2)INT{1,2}[y'dx] = (1/2)y{1,2} = (1/2)(y(2)-y(1)} = (1/2)(3/4 - 1) = -1/8

3.- dy/dx = 14 ==> y = 14x + K, K a constant ==> 9 = INT{0,3}[14x+K]dx = (7x^2 + Kx){0,3} = 7*9 + 3K = 63 + 3k ==> K = -18 and y = 14x - 18

4.- I've no the faintest idea what "state..." may possibly mean, and the second part I can't do since I don't what the functions are (y as func. of x or the other way around).

Tonio

Hello smmxwell

Originally Posted by smmxwell

The first should be pretty simple. But I'm just checking to see if I made any careless mistakes. My answer is -x/4+1/2.

Correct!

The second question seems tricky. I thought about doing it the manual way ie.e differentiating and integrating y to find it's first derivative, then divide it by 2, where I'll obtain g(x) and I' d then integrate g(x) with limits 2 and 1 respectively. Is there an easier way? Or is my idea above flawed? I don't have the answer with me at the moment.

There's no reason why you couldn't do it the way you suggest, but it's easier to say:



, as an indefinite integral

, as a definite integral



I'm sure you can finish this now.

The third should be similar. But if I integrate the first derivative to obtain it's original equation, I'd have to add a constant value to the equation. Unless I don't need to do so, I should be able to obtain y=14x, where I integrate it with limits 3 and 0. Is that correct?

You do need the constant of integration.



because when you integrate again, the ' ' term gives a term in :



Evaluate from this equation, and plug back in to find in terms of .

Regarding the fourth question, I'm not sure on how to proceed. Does the first part of the question refer to 0.5(3+8)(6)?

To be honest, I don't know what is meant by the word 'State' in the first part of this question. I presume it means, referring to the diagram above, 'Interpret the meaning of...' or 'State the value of ...'

What you need to realise is that:

• gives the area between the graph of (plotted as a function of ), the -axis and the vertical lines (ordinates) and ;

• similarly gives the area between the graph, the -axis and the horizontal lines (abscissae) and .

So in the given diagram, is the sum of the areas between the graph and the respective axes; in other words, it's the area of the rectangle bounded by the axes and the lines and .

Can you solve the question now (and yes, you will need the area of the trapezium that you've already worked out)?

Grandad

Thanks a lot guys for all the help. really appreciate it. Cheers.