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C4 integration by substitution question - where did i go wrong?

(in this question i'm using S as the integration symbol)

y = 3 / (1+4x)^1/2

Where R is the region bounded by the x axis, x = 0 and x = 2, use integration to find the area R.

let u = (1+4x)
du/dx = 4, therefore dx = 1/4 du

finding limits of u:
when x = 0, u = 1
when x = 2, u = 9

substituting back:
3 S u^-1/2 + 1/4 du - with limits 1 to 9
= 3[2u^(1/2) + x/4] limits 1 to 9
= 3[2(1+4x)^(1/2) + x/4] limits 1 to 9

substituting limits in:
3[2(1+4(9))^(1/2) + 9/4] - 3[2(1+4(1))^(1/2) + x/4]
3[14.4155...] - 3[4.7221...]
= 43.2465 -14.1664...
= 29.1 (2sf)

according to the mark scheme, the answer is 3. anyone know where i went wrong?

Reply 1

JakeyTheSnake

3[2(1+4x)^(1/2) + x/4] limits 1 to 9

Here. You don't need to substitute x back in if you've changed your limits for u.

I.e. you can either use the limits for u and what you've just integrated (generally easiest), or you can substitute x back in and use the original limits.

Reply 2

andrew.P

y= 3/(1+4x)^1/2

u = 1+ 4x

du/dx = 4, so 1/4 du = dx

so you are integrating 3/4 u^-1/2 du

then change the limits, x=2, u =9
x=0, u = 1

so integrate with limits of 9 and 1, 3/4u^-1/2 du

gives you 3/2 u^1/2 with limits 9 and 1

which equals 4.5 - 1.5

= 3

=]

Reply 3

the_somalian

JakeyTheSnake

ah ofcourse..

i think another mistake of mine was ADDING the 1/4 when i was supposed to MULTIPLY by 1/4

thanks

Reply 4

v-zero

the_somalian

ah ofcourse..

thanks

No, you're still wrong. You don't add the (1/4)du, you multiply through by it...

Reply 5

the_somalian

andrew.P

yep that works, thanks a lot

Reply 6

the_somalian

can anyone tell me how you would integrate 7/6x+2?

Reply 7

v-zero

the_somalian

can anyone tell me how you would integrate 7/6x+2?

What is the integral of 1/x ?

Reply 8

andrew.P

the_somalian

can anyone tell me how you would integrate 7/6x+2?

thats: 7 (1/6x+2)

1/6x+2 integrated becomes (1/6) ln|6x+2| (apply the ln, then work out the derivative of 6x+2 as 6.. as its integration you divide by 1/6)

so (7/6) ln|6x+2| + c is your answer

ln x

andrew.P

thats: 7 (1/6x+2)

1/6x+2 integrated becomes (1/6) ln|6x+2| (apply the ln, then work out the derivative of 6x+2 as 6.. as its integration you divide by 1/6)

so (7/6) ln|6x+2| + c is your answer

i meant 7/(6x+2)

so would it be 7 (ln|6x+2|) ?

Reply 11

Laurie161

the_somalian

Up to here you are right.

the_somalian

substituting back:
3 S u^-1/2 + 1/4 du - with limits 1 to 9

This is wrong. It is not + 1/4 du it is multiplying by 1/4 du. Originally you were multiplying by dx, you have substituted in 1/4 du for dx so you still multiply.

Also, you do not have to substitute x back in, just integrate u as you would integrate x.

You should end up with 1/4[6u^1/2] as the term you sbstitue the limits into and from this you get 3.

Reply 12

the_somalian

yep, thanks for the help laurie.. do you know if this is right?

y = 7/(6x+2)

integrated: 7(ln|6x+2|)

Reply 13

Laurie161

the_somalian

yep, thanks for the help laurie.. do you know if this is right?

y = 7/(6x+2)

integrated: 7(ln|6x+2|)

Uh don't think so if dy/dx =7/(6x+2) then y=7/6(ln l6x+2l) because when you differentiate ln of 'something' you also have to multiply by the differential of the 'something' in this case 6.

Reply 14

JakeyTheSnake

\displaystyle \int \dfrac{a}{bx + k} \ dx = \frac{a}{b} ln |bx + k| + C

Reply 15

andrew.P

the_somalian

i meant 7/(6x+2)

so would it be 7 (ln|6x+2|) ?

yeah i assumed thats what you meant that's how i did my answer:

remember you have to divide by the derivative of the (6x+2) which is 6 so 7/6 ln|6x+2|

Reply 16

ronfiles

Actually, I think you can also use this substitution:

So,
2u(du/dx)=4
lower limit = 1, upper limit = √9=3.

That way, when you integrate,
∫3 / (1+4x)^1/2 dx from 0 to 2
=∫3/u * (2u/4) du from 1 to 3
=∫3/2 du from 1 to 3
=[3u/2] 1 to 3
=9/2 - 3/2 = 6/2 = 3.

It's less of a pain, IMO, since you're getting rid of the surds - I mean, the whole point of the substitution method is to make the integration a bit simpler, right?