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    Hi, I'm really stuck on this integral problem coz i just dont know where to start, & if you could giv me a hand that would be great! thanx xx

    Q2a) Evaluate cos2x dx (with limits 90 & 0)
    b) Draw a sketch and explain your answer

    for a) i got the area = 0, but i dont know if i carried out the integration right-would cos2x integrate=sin2x or do you need to treat it like a double angle??

    I'm just completely confuzzled about the graph sorry

    Thanx for reading this
    Chrissi
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    cos2x integrates to (sin2x)/2 as it function of a function. so cos(nx) integrates to (sinnx)/n
    (coincidently this doesn't change your anwser)

    This can be used using a change of varible
    2x = t
    dt/2 = dx
    int(cos 2x dx) = int(cost dt/2) = (sint)/2 =( sin 2x )/ 2

    as for the sketch its a noraml sin graph but it "goes twice a quickly" so it just has double the osscilations
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    Thanx =) but umm, how can the area = 0?? soz, i'm just no good with these kinda questions.
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    The area doesn't equal zero, the value of the integral does. SKETCH THE GRAPH.
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    Look at the attatchment; you can see the dark blue area is cancelled out by the light blue area, hence the value of the integral is zero.
    Attached Images
     
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    i sketched the graph, & dun be so mean-i cant help if i suck at this part o higher maths. I jus dun get what it wants me 2 explain about the integral = 0 & its relationship 2 the graph. Thanx for helpin me anyways.

    Chrissi
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    it wants you to point out that, as the graph has an equal positive and negative area in that interval, you need to integrate between 0 and pi/4, then double it, to get an accurate answer. otherwise the two bits cancel out.
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    I advise you to download a program like Graphmatica at http://www8.pair.com/ksoft/.

    It'll help with those graphing and integration problems of yours.
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    omg, u are the best-thankku so much!!! =)

    Chrissi131 xx
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    thankku u guys, i actually get it now lol, thanx xx
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    I wasn't being mean, I was just emphasising the importance of sketching the graph. When you integrate a function with respect to x, any part of the function below the x-axis is considered negative area, and anything above the x-axis is positive area. Consider for example, that the velocity of a car is given by the function cos 2x. We want to know how far the car has travelled, so we integrate, because the distance is the area under the graph. So, looking at my attatchment, the dark blue represents the distance initially travelled, that is to the point corresponding to 45 deg (the bit where the graph crosses the axis). Then after 45 deg, the car starts going backwards because the velocity is negative, and covers a distance equal to the lighter blue area. If we add the displacements which are vectors, we say that at the point of 90 deg, the car has travelled zero distance - he's just drove forward, and back to the starting point, so the integral says "well, he's travelled nowhere" (hence why you get 0). But in reality he's travelled the blue and dark blue distances, and if you want to know these, you'd integrate from 0 to 45, and from 45 to 90, and add the distance ignoring any negative signs. But all you'd have to say with respect to your question "from the graph, we can see that the area above the x-axis negates the area under the x-axis".
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    lol, waste of a post then.
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    oh no, wasn't a wasted post-it actually improved ma understanding more. Thankku, after all understanding is half the battle. cheers lol! xx
 
 
 
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