# C1- Integration help!

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Can someone help me with the following two questions please

and that when , find the value of when

Right so the question comes under the topic of integration but I don't really understand what its asking me. If I was doing this in the exam I wouldn't know that I would have to integrate. So once I have integrated this. I got

What do I do next?

Next one is and the when , find the value of when . I think if I knew how to do the first question I should be able to do the second one. Any help would be appreciated. Thanks P.S how would I know that I would have to integrate this?

and that when , find the value of when

Right so the question comes under the topic of integration but I don't really understand what its asking me. If I was doing this in the exam I wouldn't know that I would have to integrate. So once I have integrated this. I got

What do I do next?

Next one is and the when , find the value of when . I think if I knew how to do the first question I should be able to do the second one. Any help would be appreciated. Thanks P.S how would I know that I would have to integrate this?

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#3

You need to recognise the question gives you dx/dt which is the the differential with respect to t of y.

So you integrate dx/dt with respect to t and you should acquire t^3+t^2+t+k. You have t^3+t^2+x which shows you do not know what the actual process of integration is doing. Rewrite dx/dt if it make it more familiar to have say dy/dx. If you don't have limits to integrate between then you must assume there is some constant (k) which can be zero. Since x had been differentiated any constant will be lost in this process when integrating we must assume there was one there.

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So you integrate dx/dt with respect to t and you should acquire t^3+t^2+t+k. You have t^3+t^2+x which shows you do not know what the actual process of integration is doing. Rewrite dx/dt if it make it more familiar to have say dy/dx. If you don't have limits to integrate between then you must assume there is some constant (k) which can be zero. Since x had been differentiated any constant will be lost in this process when integrating we must assume there was one there.

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(Original post by

when you integrate you must include a constant c

**the bear**)when you integrate you must include a constant c

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(Original post by

You need to recognise the question gives you dx/dt which is the the differential with respect to t of y.

So you integrate dx/dt with respect to t and you should acquire t^3+t^2+t+k. You have t^3+t^2+x which shows you do not know what the actual process of integration is doing. Rewrite dx/dt if it make it more familiar to have say dy/dx. If you don't have limits to integrate between then you must assume there is some constant (k) which can be zero. Since x had been differentiated any constant will be lost in this process when integrating we must assume there was one there.

Posted from TSR Mobile

**tmorrall**)You need to recognise the question gives you dx/dt which is the the differential with respect to t of y.

So you integrate dx/dt with respect to t and you should acquire t^3+t^2+t+k. You have t^3+t^2+x which shows you do not know what the actual process of integration is doing. Rewrite dx/dt if it make it more familiar to have say dy/dx. If you don't have limits to integrate between then you must assume there is some constant (k) which can be zero. Since x had been differentiated any constant will be lost in this process when integrating we must assume there was one there.

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#6

(Original post by

Right I understood that part, but what is it asking me to do when it gives values of x and t?

**Super199**)Right I understood that part, but what is it asking me to do when it gives values of x and t?

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So if you integrate dx/dt with respect to t you get the function of x so now you know x=t^3+t^2+t+k. You know a value for t when x is whatever. Sub them both in solve for your k so now you know the full equation since you know the constant k now. You can then do the next part. Note also that you have integrated with respect to t so the when you integrate the 1 it becomes t not x. If you had dy/dx is 10x+1 integrate with respect to x you get 5x^2+x+k not 5x^2+y+k

Posted from TSR Mobile

**tmorrall**)So if you integrate dx/dt with respect to t you get the function of x so now you know x=t^3+t^2+t+k. You know a value for t when x is whatever. Sub them both in solve for your k so now you know the full equation since you know the constant k now. You can then do the next part. Note also that you have integrated with respect to t so the when you integrate the 1 it becomes t not x. If you had dy/dx is 10x+1 integrate with respect to x you get 5x^2+x+k not 5x^2+y+k

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#8

2=1-1+1+c

c=1

So now you know the constant you have an equation which you know everything about: x=t^3-t^2+t+1

I hope I am not coming across too patronising. Trying to be as thorough as possible

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You know that x=t^3-t^2+t+c at the point where t=1 AND x=2 so replace both your t and your x with their corresponding values and you get 2=1^3-1^2+1+c

2=1-1+1+c

c=1

So now you know the constant you have an equation which you know everything about: x=t^3-t^2+t+1

I hope I am not coming across too patronising. Trying to be as thorough as possible

Posted from TSR Mobile

**tmorrall**)You know that x=t^3-t^2+t+c at the point where t=1 AND x=2 so replace both your t and your x with their corresponding values and you get 2=1^3-1^2+1+c

2=1-1+1+c

c=1

So now you know the constant you have an equation which you know everything about: x=t^3-t^2+t+1

I hope I am not coming across too patronising. Trying to be as thorough as possible

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#10

(Original post by

Oh wow you my friend are a lad!!!! I understand wow thanks. I got x = 7 as my answer . Thanks man!

**Super199**)Oh wow you my friend are a lad!!!! I understand wow thanks. I got x = 7 as my answer . Thanks man!

Haha yeah no problem

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(Original post by

Integration is the inverse of differentiation, so if you have a derivative and want to find a function for x in terms of t, you have to integrate it with respect to t.

To find C, substitute the values that you have been given so x=2 and t=1, so just put these values into the equation you have from integrating the derivative.

**majmuh24**)Integration is the inverse of differentiation, so if you have a derivative and want to find a function for x in terms of t, you have to integrate it with respect to t.

To find C, substitute the values that you have been given so x=2 and t=1, so just put these values into the equation you have from integrating the derivative.

a). Show that where A and B are constants to be found

B). Hench find the integral of ydx.

How do I start?

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#12

(Original post by

Help me with this one please

a). Show that where A and B are constants to be found

B). Hench find the integral of ydx.

How do I start?

**Super199**)Help me with this one please

a). Show that where A and B are constants to be found

B). Hench find the integral of ydx.

How do I start?

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