VCE Methods Question Thread!

[lzxnl]

This is what I started off doing.

Let's say I'm integrating 1/(3x+1)
Well, I know that it's going to look like ln |3x+1|
So I differentiate that by chain rule to get 3/(3x+1). I'm off by a factor of 3 -> antiderivative is 1/3 ln |3x+1| + constant

If you know what the antiderivative must look like, take a guess and then modify it. It's not a bad strategy given that ALL antiderivative rules come from differentiation.

[Maz]

thanky ou- can i ask one more question please?
how do you find the anti-derriative for (3x+2)^4?
because the rule- (ax+b)^n+1 / a(n+1) +c    isn't working on it... and how do you know when to use
this rule or another as (3x+2)^4 is in the same form as (ax+b)^n+1
the answer is 1/12*(3x+2)^4...i know the answer is right because i have checked with a few calculators and its the answer in
my textbook

[cosine]

thanky ou- can i ask one more question please?
how do you find the anti-derriative for (3x+2)^4?
because the rule- (ax+b)^n+1 / a(n+1) +c    isn't working on it... and how do you know when to use
this rule or another as (3x+2)^4 is in the same form as (ax+b)^n+1
the answer is 1/12*(3x+2)^4...i know the answer is right because i have checked with a few calculators and its the answer in
my textbook

Why won't the rule work? Remember:

So a = 3, b = 2 and n = 4

Plug these into your formula to yield:

[Maz]

because thats the wrong answer...the answer is
1/12 (3x+2)^4 and i know that answer is right because I've checked with a couple of calculators
and thats the answer in my textbook

[lzxnl]

because thats the wrong answer...the answer is
1/12 (3x+2)^4 and i know that answer is right because I've checked with a couple of calculators
and thats the answer in my textbook

If that's the right answer, you've most likely typed the question wrong. Your function is an antiderivative of (3x+2) cubed, not to the fourth power.

[Adequace]

Can someone help me with the attached? I'm stuck on what to do after my last line, I tried doing solving it from there in another book and my answer was ab/(c-a) when the answer is ab/(a-b-c).

Thanks.

Edit: My working http://imgur.com/1LtnZaG

[keltingmeith]

Can someone help me with the attached? I'm stuck on what to do after my last line, I tried doing solving it from there in another book and my answer was ab/(c-a) when the answer is ab/(a-b-c).

Thanks.

Sure. Where's the attached?

[Adequace]

Sure. Where's the attached?

Oh whoops. I reattached it to my original post, the picture of my working is too big apparently though.

[Maz]

haha- i looked at the answer to a different one- so sorry...thankyou btw

[bills]

Two questions:
1. Find the inverse function of the following:
f: R \ {3} -> R, f(x) = 1/(x-3) + 1
My answer was f^-1: R \ {1} -> R, f^-1(x) = 1/(x-1) + 3, but apparently the answer is f^-1: R \ {1} -> R, f^-1(x) = 3/(x-1) + 2. Is the book wrong?

2. A rectangular piece of cardboard has dimensions 20 cm by 36 cm. Four squares each
x cm by x cm are cut from the corners. An open box is formed by folding up the flaps.
Find a function V which gives the volume of the box in terms of x, and state the domain
for the function.

I found V = 4x(10-x)(18-x) (which is correct), but my domain is (0,10), whereas the book says it is [0,10]. Am I wrong or is the book wrong?

[brightsky]

1. Book is wrong.
2. Debatable as to who is right. I tend always to include the end points, but you could argue that if x = 0 or if x = 10, then you have no box. The distinction between the two answers is of no real significance.

[brightsky]

a/(x+a) + b/(x-b) = (a+b)/(x+c)
[a(x-b) + b(x+a)]/[(x+a)(x-b)] = (a+b)/(x+c)
(ax - ab + bx + ab)/[(x+a)(x-b)] = (a+b)/(x+c)
(a+b)x/[(x+a)(x-b)] = (a+b)/(x+c)
x/[(x+a)(x-b)] = 1/(x+c) (provided that a + b does not equal to 0)
x(x+c) = (x+a)(x-b)
x^2 + cx = x^2 - bx + ax - ab
cx = -bx + ax - ab
ab = ax - bx - cx
ab = x(a-b-c)
x = ab/(a - b - c)

[Adequace]

a/(x+a) + b/(x-b) = (a+b)/(x+c)
[a(x-b) + b(x+a)]/[(x+a)(x-b)] = (a+b)/(x+c)
(ax - ab + bx + ab)/[(x+a)(x-b)] = (a+b)/(x+c)
(a+b)x/[(x+a)(x-b)] = (a+b)/(x+c)
x/[(x+a)(x-b)] = 1/(x+c) (provided that a + b does not equal to 0)
x(x+c) = (x+a)(x-b)
x^2 + cx = x^2 - bx + ax - ab
cx = -bx + ax - ab
ab = ax - bx - cx
ab = x(a-b-c)
x = ab/(a - b - c)

ty

[Maz]

hey, can someone plz tell me how to integrate this?
21(5-7x)^3 please?
thank you in advance

[cosine]

hey, can someone plz tell me how to integrate this?
21(5-7x)^3 please?
thank you in advance

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