Integration Q. (1 Viewer)

[-pari-]

I'm having trouble finding the primitive function of things like...

(a) y' = (rt)x/ x^4 answer: y = -2/([FONT=&quot]5x^5/2[/FONT] + c

(b) y' = x^2/(rt)x answer: 2x^{[FONT=&quot]5/2[/FONT]}/ (5) + C

(c) y' = x^2 . (rt)x answer : 2x^(7/2) /7 + C

(d) y' = (3x - 5)^6 answer: [(3x - 5)^7]/21 + C

 

[bos1234]

I'm having trouble finding the primitive function of things like...

(a) y' = (rt)x/ x^4 answer: y = -2/([FONT=&quot]5x<SUP>5/2</SUP>[/FONT] + c

(b) y' = x^2/(rt)x answer: 2x<SUP>[FONT=&quot]5/2[/FONT]</SUP>/ (5) + C

(c) y' = x^2 . (rt)x answer : 2x^(7/2) /7 + C

(d) y' = (3x - 5)^6 answer: [(3x - 5)^7]/21 + C

a) can be written as : y' = (x^1/2) / x^4

a^m divide by a^n = a^(m-n)

similarly, y' = x^(1/2 - 4)
y'=x^(-7/2)

y=x^(n+1) / n+1

y=-2/5x^(-5/2) + c

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b and c are similar. You should know the index laws. Convert to indices, applky index law

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I think d is a 3 unit qn. I forgot how to do them

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thanks bye

 

[darkliight]

(d) y' = (3x - 5)^6 answer: [(3x - 5)^7]/21 + C

Just ask yourself, what differentiates to give (3x-5)^6? In general, what differentiates to give (ax+b)^c? (*)

Lets differentiate y = (ax+b)^c, see what we get and see if we can't fix it so we get what we want ...

1. y = (ax+b)^c, so chain rule (funtion inside a function rule). (#)
2. Let u = ax+b, then y = u^c.
3. Chain rule says dy/dx = dy/du * du/dx.
4. dy/du = cu^(c-1), du/dx = a.
5. dy/dx = a * cu^(c-1), but we know what u is (we invented it), so dy/dx = a * c(ax+b)^(c-1).
6. From (*) above, we wanted to get dy/dx = (ax+b)^c though, so what do we need to change in (#) to get that?
7. First thing we need is our power to be one more, so we get c as our final power, instead of c-1, so y = (ax+b)^(c+1) is a good start.
8. The stuff inside the brackets is all good, so lets leave it.
9. Now we have this extra a*c at the start, but keep in mind we just changed the c to c+1, so now we will get an a*(c+1) at the start of our derivative. How can we get rid of that? If we divide our y function by a*(c+1) we'll be good right? Because then at the end, we'll have (a*(c+1))/(a*(c+1)) which will just cancel away, so y = [ (ax+b)^(c+1) ] / (a*(c+1)) looks like the function we will need. We can add an arbitrary constant to this function too, since when we differentiate it will just differentiate away. You can check it just by differentiating y = [ (ax+b)^(c+1) ] / (a*(c+1)) + C and you should get y' = (ax+b)^c.


Now in your case, we have a = 3, b = -5, c = 6. Just plug em into that formula we just worked out, and you get:

[ (ax+b)^(c+1) ] / (a*(c+1)) + C
= [ (3x-5)^(6+1) ] / (3*(6+1)) + C
= [ (3x-5)^(7) ] / (21) + C

as required.

 

[-pari-]

ahh rite! thanksP

 

[bos1234]

y' = (3x - 5)^6

y = S (3x - 5)^6 dx

let 3x - 5 = u ..........................................du/dx = 3
.................................................................dx=du/3

y = S (u^6) dx

y = S (u^6) du/3 ...........................du/3 can be written as 1/3 x du - take 1/3 to the front since its a consant

=1/3 S (u^6) du
1/3[u^7/7]
sub back in u = 3x-5

-----------------------------------------------------------------------

\ask if anhy confusions

thanks bye

 

[darkliight]

Integration by substitution is a 3 unit topic though, and you can do this one without it. It's worth learning though if you're keen, makes these questions a lot easier.

 

[bos1234]

Integration by substitution is a 3 unit topic though, and you can do this one without it. It's worth learning though if you're keen, makes these questions a lot easier.

ya true. and you save alot of time in tests if its a hard question.

 

[-pari-]

hmm...interesting coz i found these Q's in a 2u book...

Lets differentiate y = (ax+b)^c, see what we get and see if we can't fix it so we get what we want ...

1. y = (ax+b)^c, so chain rule (funtion inside a function rule). (#)
2. Let u = ax+b, then y = u^c.
3. Chain rule says dy/dx = dy/du * du/dx.
4. dy/du = cu^(c-1), du/dx = a.
5. dy/dx = a * cu^(c-1), but we know what u is (we invented it), so dy/dx = a * c(ax+b)^(c-1).
6. From (*) above, we wanted to get dy/dx = (ax+b)^c though, so what do we need to change in (#) to get that?
7. First thing we need is our power to be one more, so we get c as our final power, instead of c-1, so y = (ax+b)^(c+1) is a good start.
8. The stuff inside the brackets is all good, so lets leave it.
9. Now we have this extra a*c at the start, but keep in mind we just changed the c to c+1, so now we will get an a*(c+1) at the start of our derivative. How can we get rid of that? If we divide our y function by a*(c+1) we'll be good right? Because then at the end, we'll have (a*(c+1))/(a*(c+1)) which will just cancel away, so y = [ (ax+b)^(c+1) ] / (a*(c+1)) looks like the function we will need. We can add an arbitrary constant to this function too, since when we differentiate it will just differentiate away. You can check it just by differentiating y = [ (ax+b)^(c+1) ] / (a*(c+1)) + C and you should get y' = (ax+b)^c.

not really sure i understand all the steps in the derivation....
is it okay just to know the formulae at the end, and sub in values? or need i know the whole procedure as well...?

 

[darkliight]

Just remembering the formula will get you the answer for those type. It's pretty easy to remember: increase the power by 1 and divide by a * (your new power).

I derived the formula so you could see that alot of these sort of problems come down to asking "what differentiates to give me this?". I just used trial and error, and changed the function until I got what I wanted. If you're stuck on something in particular, just list the lines that don't seem to make sense and i'll explain it a bit more.

 

[-pari-]

^ it's in 2u books so i'm pretty sure integrating with functions of a function is 2u.

 

[Forbidden.]

Tag cleanup ...

∫ √x/x⁴ dx = - 2 / 5x^5/2 + C

∫ x²/√x dx = 2x^5/2 / 5 + C

∫ x²√x dx = 2x^7/2 / 7 + C

∫ (3x - 5)⁶ dx = [(3x - 5)⁷] / 21 + C

...
I think d is a 3 unit qn. I forgot how to do them ...

y' = (3x - 5)^6

y = S (3x - 5)^6 dx

let 3x - 5 = u ..........................................du/dx = 3
.................................................................dx=du/3

y = S (u^6) dx

y = S (u^6) du/3 ...........................du/3 can be written as 1/3 x du - take 1/3 to the front since its a consant

=1/3 S (u^6) du
1/3[u^7/7]
sub back in u = 3x-5

-----------------------------------------------------------------------

\ask if anhy confusions

thanks bye

You mean ....

∫ (ax + b)ⁿ dx = [(ax + b)ⁿ⁺¹]/a(n+1) + C

One formula does the trick ... Not too mucha Extension 1 stuff please. These 2Uers already have enough.

 

[-pari-]

Having trouble finding the area with concave down parabolas….

1a)Find the area above the x-axis and below the parabola y = 4 – x^2 (Answer: 32/3)

1b)The line y = 3 cuts the curve y = 4 - x^2 at A (1,3) and B (-1, 3). Determine the magnitude of the region bounded by the curve y = 4- x^2 and the line y = 3. (answer: 4/3)

2) Find the area of the region enclosed between the x-axis and the curve y = 1-x^2
(answer: 4/3)

 

[SoulSearcher]

Draw the curve, makes it a lot easier to figure out what to do in these types of questions :uhhuh:

1. Basically, this is looking for the area between the curve and the x-axis. The limits are 2 and -2, and the equation to be integrated is 4-x². To find the limits, you would solve the equation 4-x² = 0.

2. The area they are looking for is between the line y = 3 and the curve. The equation to find the area would be the integral of the equation 4-x² with the limits 1 and -1, and subtracting from that integral the area of the rectangle bounded by the horizontal lines y = 9 and y = 0, and the vertical lines x = 1 and x = -1.

3. Much like the first question, except the equation is 1-x² and the limits would be 1 and -1. Limits are found by solving the equation 1-x² = 0.

 

[-pari-]

cheers SS