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Math Help - Evaluate the integral by interpreting it in terms of area

  1. #1
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    Evaluate the integral by interpreting it in terms of area

    \int_{-3}^{0} 1+ \sqrt{9-x^2} \, dx
    Last edited by mr fantastic; January 2nd 2012 at 05:41 PM. Reason: Tidied up the integral using better latex.
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    Re: Evaluate the integral by interpreting it in terms of area

    Quote Originally Posted by mathk View Post
    square root (9-x^2)^(1/2)
    If it is \int_{ - 3}^0 {\sqrt {9 - x^2 } dx} that is the area of a quarter circle.

    What circle?
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    Re: Evaluate the integral by interpreting it in terms of area

    Quote Originally Posted by mathk View Post
    \int_{-3}^0 1+ \sqrt{(9-x^2)}~ dx is the equation
    To follow the instructions in the OP
    \int_{ - 3}^0 {1 + \sqrt {9 - x^2 } dx}  = \int_{ - 3}^0 {1dx + } \int_{ - 3}^0 {\sqrt {9 - x^2 } dx}
    is the area of a rectangle plus the area of a quarter circle.
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    Re: Evaluate the integral by interpreting it in terms of area

    ok What is OP and how do you know its a rectangle and a circle?
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    Re: Evaluate the integral by interpreting it in terms of area

    Quote Originally Posted by mathk View Post
    ok What is OP
    You are told to use area to evaluate the definite integral.
    One use of definite integrals is to find area under curves.

    Quote Originally Posted by mathk View Post
    how do you know its a rectangle and a circle?
    Pre-Calculus?
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    Re: Evaluate the integral by interpreting it in terms of area

    Quote Originally Posted by Plato View Post
    You are told to use area to evaluate the definite integral.
    One use of definite integrals is to find area under curves.


    Pre-Calculus?
    umm ok
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    Re: Evaluate the integral by interpreting it in terms of area

    I have no idea what "umm ok" means!

    The problem asked you to evaluate this by interpreting it as an area so you should have thought about the geometry determined by these functions' graphs. The integral can be interpreted as "the area under the graph". Your integrand was 1+ \sqrt{9- x^2.

    You should, perhaps from pre-calclulus, perhaps from Cartesian geometry, recognise that the graph of y= 1 is a horizontal straight line and, from x= -3 to 0, from y= 0 to y= 1, is a rectangle with length 0-(-3)= 3 and height 1. What is the area of that rectangle?

    You should also be able to recognixze that y= \sqrt{9 -x^2} is the part of the circle x^2+ y^2= 9 above the x-axis (because the square root is positive). That circle has center (0, 0) and radius 3 and so goes, along the x-axis, from -3 to 3. Going from -3 to 0 gives you half of that circle and taking y positive gives you half of that- one quarter of the circle. What is that area?
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    Re: Evaluate the integral by interpreting it in terms of area

    Quote Originally Posted by mathk View Post
    What? I dont get it
    The geometric interpretation is the easiest method, as has been pointed out to you. But if you were going to use the substitution, let \displaystyle \begin{align*} x = 3\sin{\theta} \implies dx = 3\cos{\theta}\,d\theta \end{align*}, and note that when \displaystyle \begin{align*} x = -3, \theta = -\frac{\pi}{2} \end{align*} and when \displaystyle \begin{align*} x = 0, \theta = 0 \end{align*}, then the integral becomes...

    \displaystyle \begin{align*} \int_{-3}^0{ 1 + \sqrt{ 9 - x^2 } \, dx } &= \int_{-3}^0{1\,dx} + \int_{-3}^0{\sqrt{9-x^2}\,dx} \\ &= \left[ x \right]_{-3}^0 + \int_{ -\frac{ \pi }{ 2 } }^0{ \sqrt{ 9 - \left( 3\sin{\theta} \right)^2 } \, 3\cos{\theta} \,d\theta } \\ &= 0 - (-3) + 3\int_{-\frac{\pi}{2}}^0{ \sqrt{9 - 9\sin^2{\theta}} \, \cos{\theta}\,d\theta } \\ &= 3 +  3\int_{-\frac{\pi}{2}}^0{\sqrt{9\left( 1 - \sin^2{\theta} \right)} \, \cos{\theta}\,d\theta} \\ &= 3 + 3 \int_{-\frac{\pi}{2}}^0{ \sqrt{ 9\cos^2{\theta} } \,\cos{\theta}\,d\theta} \\ &= 3 + 3\int_{-\frac{\pi}{2}}^0{ 3 \cos{\theta} \cos{\theta}\,d\theta} \\ &= 3 + 9\int_{-\frac{\pi}{2}}^0{\cos^2{\theta}\,d\theta} \\ &= 3 + 9\int_{-\frac{\pi}{2}}^0{\frac{1}{2} + \frac{1}{2}\cos{2\theta}\,d\theta} \\ &= 3 + 9\left[ \frac{1}{2}\theta + \frac{1}{4}\sin{2\theta} \right]_{-\frac{\pi}{2}}^0 \end{align*}

    \displaystyle \begin{align*} &= 3 + 9\left\{ \left[ \frac{1}{2}\cdot 0 + \frac{1}{4}\sin{(2\cdot 0)} \right] - \left[ \frac{1}{2}\left(-\frac{\pi}{2}\right) + \frac{1}{4}\sin{ \left[2 \left( -\frac{\pi}{2} \right) \right] } \right] \right\} \\ &= 3 + \frac{9\pi}{4} \end{align*}

    which is the same as the area of the rectangle of length 3 units and width 1 unit plus the area of the quarter circle of radius 3 units, as you were advised
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    Re: Evaluate the integral by interpreting it in terms of area

    @Prove It, Thank you for your contribution to this question.
    However, I must say that I think it violates the sprit of the question.
    I think that your approach is a least twenty years out of date.
    That is, I think the author of the question is spot on to the current state in mathematics education. I have a firmly held thought that within twenty years no mainstream calculus text will contain a chapter on “techniques of integration”. The wide availability of computer algebra systems have made those topics obsolete for the most part. (Of course they will be great topics for Mathematics Education courses and courses in the History of Mathematics.)
    Look at this link to see what I mean.
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    Re: Evaluate the integral by interpreting it in terms of area

    Quote Originally Posted by Plato View Post
    @Prove It, Thank you for your contribution to this question.
    However, I must say that I think it violates the sprit of the question.
    I think that your approach is a least twenty years out of date.
    That is, I think the author of the question is spot on to the current state in mathematics education. I have a firmly held thought that within twenty years no mainstream calculus text will contain a chapter on “techniques of integration”. The wide availability of computer algebra systems have made those topics obsolete for the most part. (Of course they will be great topics for Mathematics Education courses and courses in the History of Mathematics.)
    Look at this link to see what I mean.
    Violates the spirit of the question, perhaps.
    Obsolete, hardly. There are many integrals that would require trigonometric/hyperbolic substitution that can't be solved using a geometric interpretation. And the use of a CAS in those situations is just plain lazy and irresponsible. How would they be programmed in future if everyone thought the method of solution was "obsolete"?

    Besides, my post was in response to the original poster asking for clarification when someone else suggested using this method. I pointed out that the geometric interpretation is the quicker and better method.
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    Re: Evaluate the integral by interpreting it in terms of area

    Quote Originally Posted by Plato View Post
    @Prove It, Thank you for your contribution to this question...

    I think that your approach is a least twenty years out of date.
    That is, I think the author of the question is spot on to the current state in mathematics education. I have a firmly held thought that within twenty years no mainstream calculus text will contain a chapter on “techniques of integration”. The wide availability of computer algebra systems have made those topics obsolete for the most part. (Of course they will be great topics for Mathematics Education courses and courses in the History of Mathematics.)
    Look at this link to see what I mean.
    I think in what passes for the real world they have been out of date for far longer. In fields that actually use such integrals it has been virtually universal practice to look them up in tables since before I started work outside of a university in the late 70's. The tables by Gradshteyn & Ryzhik have been the standard since the mid 60's, when they replaced tables going back at least as far as the 1820's.

    CB
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