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Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Find the area underneath the curve.?

Let R be the region between the curve

y = 3(x^2)-2/(x^2)

and the x-axis, for x between 1 and 2.

Options:

A. 3

B. 6

C. 5

D. 8

------

I know how to set up the problem, I just get stuck with derivatives for fractions.

Any help is appreciated. Thank you.

2 Answers

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  • JG
    Lv 5
    1 decade ago
    Favorite Answer

    I'll assume you didn't leave out parenthesis so will add spaces and remove them as they are unneeded for clarity:

    y = 3x^2 - 2/x^2

    the integral is

    x^3 + 2/x

    evaluate this at 2 and 1 and subtract

    (2^3 + 2/2) - (1^3 + 2/1)

    9 - 3 = 6

  • mountainpenguin
    Lv 4
    1 decade ago

    integrate the function with respect to x between 1 and 2:

    {2,1}∫3x² - 2/x² dx = {2,1}[x³ + 2/x] = (2³ + 2/2 - 1³ -2/1) = 8 + 1 - 1 - 2 = 6

    so B

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