Help I Cant Figure Out How To Do These Q!!! (1 Viewer)

[Cattle]

Can anyone please help me i have a bit of an idea with these questions but im not to sure...

1) The gradient of a curve is given by dy/dx 2x-3. what is the equation of the curve if:
a) It passes through the origin?
b) It passes through (1,-2)
c) It passes through (1,2)

With the next q i can do part a) but i dont get the answer that is used in b) can someone please help!
2) a) differentiate 2x^2 +1(all over) 3x^2- 4
b) Hence evaluate integration from 0-1 (cant figure out how to do the sing thingy on here) x (over) (3x^2- 4)^2

3) Find the area between y= 1/x^2 ( 1 over x^2), the y axis and the lines y=1, y=2. Express the answer in surd form...... i can do the first bit of it but dont understand how to put it in surd form????

4) Find the total area of the region bounded by f(x)= x^3 - x and g(x)= 1 - x^2..... i know you have to find the point of intersection but i cant figure out how to do this

Any ideas would be greatly appreciated

 

[word.]

Hi Cattle.

1) The gradient of a curve is given by dy/dx 2x-3. what is the equation of the curve if:
a) It passes through the origin?
b) It passes through (1,-2)
c) It passes through (1,2)

If the gradient function of the curve is given by 2x - 3, then the function is given by x² - 3x + C (Integrating).
i.e. y = x² - 3x + C

a) At (0,0), y = 0 when x = 0,
so 0 = 0 - 0 + C, C = 0
hence the equation of the curve is given by f(x) = x² - 3x.
It works in a similar fashion for b) and c).


With the next q i can do part a) but i dont get the answer that is used in b) can someone please help!
2) a) differentiate 2x^2 +1(all over) 3x^2- 4
b) Hence evaluate integration from 0-1 (cant figure out how to do the sing thingy on here) x (over) (3x^2- 4)^2


Differentiating we get: [4x(3x² - 4) - 6x(2x² + 1)]/(3x² - 4)²
= -22x/(3x² - 4)²

i.e. d/dx (2x² +1)/(3x² - 4) = -22x/(3x² - 4)²
so INT. -22x/(3x² - 4)² = (2x² +1)/(3x² - 4) + C
and so INT. x/(3x² - 4)² = -(2x² +1)/22(3x² - 4) (multiplying both sides by -1/22)
then sub in as per usual.


3) Find the area between y= 1/x^2 ( 1 over x^2), the y axis and the lines y=1, y=2. Express the answer in surd form...... i can do the first bit of it but dont understand how to put it in surd form????

The question asks you to put it into surd form because you will have to convert it into index form to integrate it... (unless you can see it straight off of course).

Surd form is basically the roots to nth power.

If you have x^3/2 you know this is equivalent to Sqrt[x³]

x² = 1/y
x = 1/Sqrt[y] = y^-1/2
Integrating we get: 2y^1/2 + C = 2*Sqrt[y] + C


4) Find the total area of the region bounded by f(x)= x^3 - x and g(x)= 1 - x^2..... i know you have to find the point of intersection but i cant figure out how to do this

x³ - x = 1 - x²
x³ + x² - x - 1
This is basic trinomial factorisation you should practise:
x³ + x² - x - 1 = x(x² - 1) + 1(x² - 1)
= (x - 1)(x + 1)(x + 1)
Intersection points occur at x = +-1

 

[Riviet]

Hey there, Cattle. ^^

Q4
Since the function f(x)=x³-x is odd, ie f(-x)=-f(x)
therefore the area from -1 to 0 = area from 0 to -1 [we were asked to find the area, not evaluate the integral, therefore we ignore negative area rules]
Now if you sketch both of the graphs on the same number plane, you will see that we are finding the area under the parabola from -1 to 1, minus the area from -1 to 0 of f(x)=x³-x, plus the area of f(x)=x³-x from 0 to 1.
Therefore we can take the area from 0 to 1 of f(x)=x³-x, cut it, rotate it 180 degrees, and stick it where we minused the area under f(x)=x³-x from -1 to 0.
Hence, we end up with the area under the parabola g(x) which is equivalent to the area bounded by f(x) and g(x).
.: the total area of the region bounded by f(x)= x^3 - x and g(x)= 1 - x^2 is:
_-1∫¹ 1-x² dx]
= 2[₀∫¹ 1-x² dx , since the area under a curve taken from both sides equidistant to the axis of the even function is equal.
= 2[x-x³/3]¹₀
= 2(1-1/3-0)
= 4/3 units²

 

[Cattle]

Differentiating we get: [4x(3x² - 4) - 6x(2x² + 1)]/(3x² - 4)²
= -22x/(3x² - 4)²

i.e. d/dx (2x² +1)/(3x² - 4) = -22x/(3x² - 4)²
so INT. -22x/(3x² - 4)² = (2x² +1)/(3x² - 4) + C
and so INT. x/(3x² - 4)² = -(2x² +1)/22(3x² - 4) (multiplying both sides by -1/22)
then sub in as per usual.

I dont understand how you did this can you please emplain.....i understand the differentiation part but i dont know how you got the answer that you did when you integrated and why you are multiplying by -1/22....please help!!!

 

[gman03]

I dont understand how you did this can you please emplain.....i understand the differentiation part but i dont know how you got the answer that you did when you integrated and why you are multiplying by -1/22....please help!!!

Recall that integration canbe refer as anti-derivative, meaning that it can almost reverse the differentiation.

if we differentiate a function f(x), we get d/dx f(x).

That means the anti-derivative of d/dx f(x) is f(x) + C.
So the integral of d/dx f(x) is f(x) + C.


In this question

f(x) = [4x(3x² - 4) - 6x(2x² + 1)]/(3x2 - 4)²

and you said u understand why

d/dx f(x) = -22x/(3x² - 4)²


The question ask you to find the integral of x/(3x² - 4)²

but

x/(3x² - 4)² = -1/22 * -22x/(3x² - 4)²
= -1/22 * d/dx f(x)

so integral of that is same as integral of -1/22 * d/dx f(x), which is -1/22 f(x) + C

 

[Cattle]

ok cool thankyou