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Thread: integrals and area

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    integrals and area

    whats the area between:
    y=underroot(2x), y axis and the line y=4

    ans:32/3
    and area between x^2=4ay and its latus rectum
    Last edited by sssona09; 12 Feb 2018 at 5:19 PM.

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    Re: integrals and area

    You can get the upper bound for the first integral by solving for . Then evaluate the integral and subtract it from a rectangle to get your answer.

    Alternatively, you can get the answer directly by switching and , rearranging to make the subject and integrating from 0 to 4. The curve will become a parabola, but it doesn't affect the integral.
    ______________________________________________

    The latus rectum of is .

    If you shift the parabola down by , we can integrate on the x-axis (the areas will still be the same). To get the bounds, find the x-intercepts of the new shifted parabola. Finally, take the absolute value of the integral to get the area.
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

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