Year 13 Maths Help Thread

Integration by substitution question. (See attached working and correct answer - ques c).

The answer in the book has 2 fractions but i only got one. Help :s-smilie:

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Reply 1182

Original post by kiiten

Integration by substitution question. (See attached working and correct answer - ques c).

The answer in the book has 2 fractions but i only got one. Help :s-smilie:

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Because you integrated with respect to u whilst keeping an x in there.

Reply 1183

Original post by SeanFM

Because you integrated with respect to u whilst keeping an x in there.

Ahh of course, but how do i get rid of the x as it doesnt cancel out?

Reply 1184

Original post by kiiten

Ahh of course, but how do i get rid of the x as it doesnt cancel out?

The method you are using is Integration by substitution :smile:

Reply 1185

Original post by SeanFM

The method you are using is Integration by substitution :smile:

So where u = 2x+1 do i rearrange to get x?

(u - 1) / 2 = x

then integrate that?

Reply 1186

Original post by kiiten

So where u = 2x+1 do i rearrange to get x?

(u - 1) / 2 = x

then integrate that?

You tell me :smile:

If you are stuck in a problem in maths or in life it helps to try things and explore different options to see where it would get you. :rofl:

Reply 1187

Original post by SeanFM

You tell me :smile:

If you are stuck in a problem in maths or in life it helps to try things and explore different options to see where it would get you. :rofl:

Unless ive done something wrong now i end up with:

( 1/2u^2 - 1 )(1/6u^6) :s-smilie:

Reply 1188

Original post by kiiten

Unless ive done something wrong now i end up with:

( 1/2u^2 - 1 )(1/6u^6) :s-smilie:

Then you have gone wrong, and I don't know where as you have just that line and u^-6 is in there.

Go back to your original working just before where you integrated x and apply the substitution.

Reply 1189

Original post by SeanFM

Then you have gone wrong, and I don't know where as you have just that line and u^-6 is in there.

Go back to your original working just before where you integrated x and apply the substitution.

What am i doing wrong - i keep getting different answers every time :3

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Original post by kiiten

What am i doing wrong - i keep getting different answers every time :3

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If

u=2x+1

then

2x=u-1 \Rightarrow 4x=2(u-1)

In the integral,

\frac{4x}{2}

simply becomes

(u-1)

You made a mistake by writing

\frac{4u-1}{2}

because that's not the correct substitution.

If you want, you can get

x=\frac{u-1}{2}

which then means that

4x=\frac{4u-4}{2}

so that -1 should be -4.

Reply 1191

Original post by RDKGames

If

u=2x+1

then

2x=u-1 \Rightarrow 4x=2(u-1)

In the integral,

\frac{4x}{2}

simply becomes

(u-1)

You made a mistake by writing

\frac{4u-1}{2}

because that's not the correct substitution.

If you want, you can get

x=\frac{u-1}{2}

which then means that

4x=\frac{4u-4}{2}

so that -1 should be -4.

Ohh yeah ugh i forgot to multiply the 1 and 4. Ok ive done that but now i get this D:

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Original post by kiiten

Ohh yeah ugh i forgot to multiply the 1 and 4. Ok ive done that but now i get this D:

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You can't integrate like that, it doesn't make sense.

Just expand the brackets inside the integral and then integrate each term...

Reply 1193

Original post by RDKGames

You can't integrate like that, it doesn't make sense.

Just expand the brackets inside the integral and then integrate each term...

Oh so the third line would be the integral of u^6 - u^5 ?

Original post by kiiten

Oh so the third line would be the integral of u^6 - u^5 ?

Yes.

Reply 1195

Original post by RDKGames

Yes.

Thanks

Reply 1196

jamestg

Edexcel C3 June 2006 4d

Can you say something like:

"You would have -5 on the LHS and when taking natural logs of both sides it is not possible to take the natural log of a negative, therefore T has to be greater than 25"

Original post by jamestg

Edexcel C3 June 2006 4d

Can you say something like:

"You would have -5 on the LHS and when taking natural logs of both sides it is not possible to take the natural log of a negative, therefore T has to be greater than 25"

Post the question or link it.

Reply 1198

username1162972