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Math Help - definite integrals

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    definite integrals

    Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals



    Can somebody please help me with this one, I am not understanding how to approach it. I know it forms a fourth of a circle because of the bounds but that is all I am understanding.
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    Re: definite integrals

    Quote Originally Posted by sonalig View Post
    Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals .

    That is simply the area enclosed by the unit circle in the first quadrant.
    How much is that?

    Look at this page
    Last edited by Plato; April 21st 2013 at 02:49 PM.
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    Re: definite integrals

    Let's think of x and 1 in the given question as two sides of a triangle like in this image below:

    definite integrals-triangle.png

    \cos{(\theta)} = \frac{x}{1} = x

    \sqrt{1-x^2} = \sqrt{1-\cos^2{(\theta)}} = \sin{(\theta)}

    \text{And } dx = -\sin{(\theta)}\,\,d\theta

    Now for lower limit x = 0:
    x = \cos{(\theta)} = 0 \therefore \theta = \frac{\pi}{2}


    And for upper limit x = 1:

    x = \cos{(\theta)} = 1 \therefore\,\, \theta = 0


    Back to the main question which can be written as:

    \begin{align*}\int_0^1 \sqrt{1-x^2}\,\,dx =& -\int_{\frac{\pi}{2}}^0 \sin^2{(\theta)}\,\,d\theta \\=& \int_{\frac{\pi}{2}}^0 \frac{\cos{(2\theta)}-1}{2}\,\,d\theta....\text{[Using trigonometry]} \\=& \frac{\sin{(2\theta)}}{4} -\frac{\theta}{2}\Bigg]_{\frac{\pi}{2}}^0 \\=& 0 - 0 - \frac{\sin{(\pi)}}{4} + \frac{\pi}{4} \\=& \frac{\pi}{4}....\text{[for 1st quadrant]}\end{align*}
    Last edited by x3bnm; April 21st 2013 at 05:45 PM.
    Thanks from agentmulder
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    Re: definite integrals

    Quote Originally Posted by x3bnm View Post
    Let's think of x and 1 in the given question as two sides of a triangle like in this image below:
    Click image for larger version. 

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    \cos{(\theta)} = \frac{x}{1} = x
    \sqrt{1-x^2} = \sqrt{1-\cos^2{(\theta)}} = \sin{(\theta)}
    \text{And } dx = -\sin{(\theta)}\,\,d\theta
    Now for lower limit x = 0:
    x = \cos{(\theta)} = 0 \therefore \theta = \frac{\pi}{2}
    And for upper limit x = 1:
    x = \cos{(\theta)} = 1 \therefore\,\, \theta = 0
    Back to the main question which can be written as:
    \begin{align*}\int_0^1 \sqrt{1-x^2}\,\,dx =& -\int_{\frac{\pi}{2}}^0 \sin^2{(\theta)}\,\,d\theta \\=& \int_{\frac{\pi}{2}}^0 \frac{\cos{(2\theta)}-1}{2}\,\,d\theta....\text{[Using trigonometry]} \\=& \frac{\sin{(2\theta)}}{4} -\frac{\theta}{2}\Bigg]_{\frac{\pi}{2}}^0 \\=& 0 - 0 - \frac{\sin{(\pi)}}{4} + \frac{\pi}{4} \\=& \frac{\pi}{4}....\text{[for 1st quadrant]}\end{align*}
    @x3bnm
    Why would anyone changed such a transparently simple question into such a complicated mess?
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    Senior Member x3bnm's Avatar
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    Re: definite integrals

    I guess I like complicated things . Can you tell me how you would solve it if you don't mind telling me?
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    Re: definite integrals

    Quote Originally Posted by sonalig View Post
    Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals



    Can somebody please help me with this one, I am not understanding how to approach it. I know it forms a fourth of a circle because of the bounds but that is all I am understanding.
    Surely you know that the area of a circle is calculated by \displaystyle A = \pi \, r^2 . Here you have a quarter of the circle of radius 1, so what is the area?
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    Re: definite integrals

    IMHO x3bnm has shown very nicely how to work integrals using trigonometric substitutions. The OP asked for sums and differences of areas which x3bnm provided using simple geometric formulas of right triangles.

    Last edited by agentmulder; April 22nd 2013 at 02:54 AM.
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    Re: definite integrals

    Quote Originally Posted by agentmulder View Post
    IMHO x3bnm has shown very nicely how to work integrals using trigonometric substitutions. The OP asked for sums and differences of areas which x3bnm provided using simple geometric formulas of right triangles.

    Surely an integral with trigonometric substitutions is much more difficult than \displaystyle \frac{\pi \, r^2}{4}. It's 100% correct and beautifully executed (and LaTeXed), but the question asked to use simple geometric formulas. Surely there are few more simple than the formula for the area of a circle...
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    Re: definite integrals

    We know that the function of the given question is:

    y = \sqrt{1-x^2}........\text{(1)}

    By polar coordinate we know that x = r\cos(\theta) and y = r\sin(\theta)
    If we plugin these values into equation (1) we get:

    \begin{align*}r\sin{(\theta)} =& \sqrt{1-r^{2}\cos^2{(\theta)}} \\ r^2\sin^2{(\theta)} =& 1 - r^{2}\cos^2{(\theta)}\\ r^2(\sin^2{(\theta)} + \cos^2{(\theta)}) =& 1\\ \therefore\,\, r =& 1\end{align*}

    For upper limit x = 1
    x = r\cos{(\theta)} = \cos{(\theta)} = 1 \therefore\,\, \theta = 0

    For lower limit x = 0

    x = r\cos{(\theta)} = \cos{(\theta)} = 0 \therefore\,\, \theta = \frac{\pi}{2}

    So the area is one fourth of the full quadrant.

    Area = \pi r^2 - \frac{3 \pi r^2}{4} = \frac{\pi}{4}.........\text{[because  } r = 1 \text{]}
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