Integration AS P1

https://gyazo.com/684ad90ff737225cc1415511fec86551

Some help doing part one of the question please. I think i can do part 2 and part 3 on my own but I can't get the expression right. Thanks

Original post by Jian17

https://gyazo.com/684ad90ff737225cc1415511fec86551

Some help doing part one of the question please. I think i can do part 2 and part 3 on my own but I can't get the expression right. Thanks

You can integrate

\sqrt{4-x}

by substitution, if you so must.

u=4-x

Reply 2

Angel_Chen

20

Original post by Jian17

https://gyazo.com/684ad90ff737225cc1415511fec86551

Some help doing part one of the question please. I think i can do part 2 and part 3 on my own but I can't get the expression right. Thanks

x^1/2 is the same as sqr x.

Reply 3

Jian17

OP

10

How do i find the second expression ? I don't understand that sign and the y after it.

Original post by Jian17

How do i find the second expression ? I don't understand that sign and the y after it.

What sign? Explain clearly what's confusing you.

Reply 5

Jian17

OP

10

These https://gyazo.com/5a6d329b3f4b0c3fac34b74d0e1322ae

I got this answer by the way:

8x - ((4-x)^3/2 / 3/2)

Reply 6

Jian17

OP

10

https://gyazo.com/2336a6bb53201cbfeafaaa3c882b4013

These are the answers, why does he multiply and divide by -1 ?

Original post by Jian17

These https://gyazo.com/5a6d329b3f4b0c3fac34b74d0e1322ae

I got this answer by the way:

8x - ((4-x)^3/2 / 3/2)

Not sure how you don't understand the basic notation for differentiation and integration...

\dfrac{dy}{dx}

means the derivative of

y

w.r.t

x

. So differentiate

8-\sqrt{4-x}

w.r.t

x

\displaystyle \int y .dx

means integral of

y

w.r.t

x

. So integrate

8-\sqrt{4-x}

w.r.t

x

That's almost the answer for the integration part, except your answer differentiates to

8+(4-x)^{1/2}

instead of

8-(4-x)^{1/2}

Reply 8

Acc Dos

6

Original post by RDKGames

You can integrate

\sqrt{4-x}

by substitution, if you so must.

u=4-x

There's no point in complicating things by using U substitution in this case. Just use the reverse chain rule, it's an easy expression.

Add one to the power, then divide the whole function by the new power and also the derivative of the inner function.

Original post by Acc Dos

There's no point in complicating things by using U substitution in this case. Just use the reverse chain rule, it's an easy expression.

Add one to the power, then divide the whole function by the new power and also the derivative of the inner function.

Reverse chain rule comes FROM substitution for linear expressions in the first place. It is not overxomplicating anything whatsoever.
If OP doesn’t understand this first, then whats the point in just blindly applying reverse chain rule if they dont know where it comes from/why it works.

Reply 10

Acc Dos

6

I know that it does, that's why I specified 'U substitution' didn't really know how to word that, my bad. But I guess you're right about OP needing to understand some fundamentals beforehand.I was actually taught the reverse chain rule before I learnt how to do that kind of substitution, strange now that I think about it.

Integration AS P1

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