What is the derivative, anti-derivative, and integral of e^(r/2)?

Also what is e^(r/2)dr?

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  • 1 decade ago
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    Derivative of e^(r/2) with respect to r

    = d(e^(r/2))/dr

    =(d(e^(r/2))/d(r/2)) x (d(r/2)/dr)

    =e^(r/2)x(1/2)

    =(e^(r/2))/2

    Anti-derivative or indefinite integral of e^(r/2):

    = ∫(e^(r/2))dr

    = (e^(r/2))/(1/2)

    = 2e^(r/2)

    To find the integral (Conventionally, the Definite integral), you need the lower and upper limits of integration. That has not been stated in this question. Normally, if the limits are not stated, the integral means the indefinite integral and is equal to the anti-derivative.

    Oh, and e^(r/2)dr actually means are under the graph of the function e^(r/2) for a very very small chance in r (that is known as dr). Look at the image on this web-page for example: http://en.wikipedia.org/wiki/File:Integral_approxi...

    The length of the green rectangles maybe treated as dr.

    Source(s): It's all there in my head, of course! :)
  • 4 years ago

    the subject as suggested seems to be f '' (x) no longer a single spinoff. you are going to be able to desire to combine two times. it is the reason they gave 2 f(x) values. f ''(x) = 20x3 + 12x2 + 4 f ' (x) = (20/4)x^4 + (12/3)x^3 + 4x + C f ' (x) = 5x^4 + 4x^3 + 4x + C f(x) = (5/5)x^5 + (4/4)x^4 + (4/2)x^2 + Cx + D f(x) = x^5 + x^4 + 2x^2 + Cx + D now initiate with f(0) = 7 f(x) = 0.5 + 0^4 + 2*(0^2) + C(0) + D = 7 so D=7 now f(a million) = 9 f(x) = a million^5 + a million^4 + 2*(a million^2) + C(a million) + 7 = 9 f(x) = a million+a million+2+C+7=9 C=-2 f(x) = x^5 + x^4 + 2x^2 -2x + 7 dnadan1

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