Integration - pre-exam preperation (1 Viewer)

[-pari-]

1) ∫ (rt)x + (1/x^2) where limits are a = 1, b = 5,

2) find the area bound by the curve y = (4 – x^2)^1/2 the x axis and the y axis in the first quadrant.
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3) find the area enclosed between the curve y = x^3 – x^2 and the y axis between y = -1 and y = 25
<O
4) find the exact area enclosed between the curve y = (4 – x^2)^1/2 and the line x – y +2 = 0
Answer (pi – 2)
<O</O
5 ) ok I’m having one major problem – figuring out where and when to use absolute values when integrating.
You know when sometimes the curve is in the negative quadrant? So you put absolute values when integrating, to indicate that area is always positive.
So like…eg the are bound by the curve y = (x + 3) ^ 2 the y axis and the lines y = 9, y = 16. find the volume of solid formed if this area is rotated about the x axis.
When you integrate….you say ∫f(x) dx = F(b) – F(a) where F(x) is the primitive function of f(x)
But say the F(b) is the negative one…..do you calculate it like
|F(b)| - F(a)
Or | F(b) – F(a)|
I don’t get it.
<O</O
6) the area bound by the parabola y = 2x – x^2 the y axis and the line y = 1 is rotated about the x axis, find the volume generated.
Since its volume, its v = (pi)∫ y^2 dx
And since it’s around the y axis, you convert the subject to X …in this case since of all x can’t be shifted to the one side, it is okay to say x^2 = y – 2x and integrate that since in any case we need to square it. But suppose the question asked for area instead of volume….how would I do that?
<O</O
7) show that the tangent at the point P where x = 3 on the curve y = e^x has the equation xe^3 – y = 2e^3 . this tangent meets the x axis at Q. if R is the foot of the perpendicular from P on the x axis, find the coordinates of Q and the length of QR.
(don’t know how to do this second bit)
<O</O
8) for what values of x is the curve e^(x^2 – 2x + 3) monotonic increasing?
<O>with this one, you differentiate the equation then equate x to 0. but what i dont get is in doing so, you elimate e^x by 'taking it to the other side'....but the power of e is x as well! how can u just get rid of it??? (does that make sense?) </O>
<O</O
9) evaluate ∫xe^x where the limits are a = 0, b= 1
I’ve been told that you first differentiate this and then work backwards….is that right? Is there any other method I’d be better off using?
<O</O
<O any help would be oh-so much appreciated!!! </O>

 

[jemsta]

1) int rt(x) is 2/3x^3/2 and int of 1/x^2 is -1/x
then just work from there

9) the best way to tackle this question is by integration by parts, which adv maths does not have to learn.
is the answer to question nine 1? if it is then ill show you that way on how to do it.

 

[pLuvia]

2. Just a quarter of a circle, radius 2
3. Find where x equals those y values, do a quick sketch to see where the graph is negative then find the area making sure you make the negative area positive
4. A quarter of the circle in the 2nd quadrant minus the triangle the linear equation forms with the x and y axis
5. Where the graph is below the x axis you make that area positive, so you put the absolute values in.
int.f(x)=F(b)-F(a) is just the area between x=a,b. If in those limits the graph is below the x axis, put absolute values in. If F(b) is negative it does not necessarily mean the area is all negative, and besides F(b) isn't an area it's just one of values.
6. Actually the volume would be V=pi int. f(y)²dy, so you would make x the subject by completing the square then you'll get y as a function of x. And no you wouldn't be able to do it that way you said since there is an x on one side, but you are integrating with respect to y
7. R is the point if a line is drawn from P down to the x axis so the point R is (3,0), then basically let y=0 for the tangent to find the coordinates of Q then use the abscissa of Q minus the abscissa of R
8. Differentiate it once, (2x-2)e^x2-2x+3>0, well e^x can never be equal to zero for any x value so you just say that e can never be zero, so you solve for 2x-2, so x>1 is monotonic increasing
9. You'll need to use integration by part for this a 4 unit method. Basically the product rule but then you integrate it
d/dx(uv)=u'v+uv'
uv=int. u'v dx+int. uv' dx
Integration by part formula
int.uv' dx=uv-int. u'v dx

Or you could probably just differentiate it then you'll see you have an xe^x then you integrate both sides and you should get your answer