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Homework Help: Evaluate this integral please

  1. Dec 12, 2008 #1
    Hi,

    Can anyone tell me if the following of my integration is right or wrong?

    [tex]
    \int_{-2}^{2} \sqrt{4 - x^2} \ dx
    [/tex]

    let

    [tex]x = 2 sin \theta[/tex]

    then

    [tex]\frac{dx}{d\theta}=2cos\theta, \ dx=2cos\theta \ d\theta[/tex]

    also

    [tex]4-x^2 = 4-(2sin\theta)^2
    = 4 - 4sin^2\theta = 4(1-sin^2\theta)=4cos^2\theta
    [/tex]

    thus

    [tex]
    \sqrt{4-x^2}=\sqrt{4cos^2\theta}
    [/tex]

    [tex]\int_{-2}^{2} \sqrt{4 - x^2} \ dx[/tex]
    [tex]= \int_{-2}^{2} \sqrt{4cos^2\theta} \ (2cos\theta \ d\theta)[/tex]
    [tex]= \int_{-2}^{2} 2cos\theta \ (2cos\theta \ d\theta)[/tex]
    [tex]= 4\int_{-2}^{2} cos^2\theta \ d\theta[/tex]
    [tex]= 4\int_{-2}^{2} \frac{1+cos2\theta}{2} \ d\theta[/tex]
    [tex]= \frac{4}{2} \int_{-2}^{2} (1+cos2\theta) \ d\theta[/tex]

    let [tex]u=2\theta[/tex] then [tex]\frac{du}{d\theta}=2, \ d\theta = \frac{1}{2}du[/tex]

    [tex]= 2 \int_{-2}^{2} (1+cos2\theta) \ d\theta[/tex]
    [tex]= 2 \int_{-2}^{2} (1+cos \ u) \ (\frac{1}{2}du)[/tex]
    [tex]= \frac{2}{2}\int_{-2}^{2} (1+cos \ u) \ du[/tex]
    [tex]= \int_{-2}^{2} (1+cos \ u) \ du[/tex]

    [tex]= u + sin \ u]^{2}_{-2}[/tex]

    [tex]= 2\theta + sin2\theta]^{2}_{-2}[/tex]

    [tex]= [2(2) + sin2(2)] - [2(-2) + sin2(-2)][/tex]

    [tex]= (4 + sin4) - [-4 + sin(-4)][/tex]

    [tex]= 4 + sin4 + 4 - sin(-4)[/tex]

    [tex]= 8 + sin4 - sin(-4)[/tex]

    [tex]= 8 + 0.07 - (-0.035)[/tex]

    [tex]= 8 + 0.07 + 0.035[/tex]

    [tex]= 8 + 0.07 + 0.035[/tex]

    [tex]= 8.105[/tex]

    Thenk you very much for evaluating
     
  2. jcsd
  3. Dec 12, 2008 #2
    Your integration is wrong, but no worries.

    First of all, notice that your integrand is an even function, which means that you can change your integral to [tex]2\int_{0}^{2} \sqrt{4 - x^2} \ dx[/tex] to make things (slightly) easier.

    However, the crucial mistake you made is forgetting to change the limits of integration when you switched the variables x and theta. The choice of substitution is fine and your algebra also looks good. Just remember x = 2sin(theta) => theta = arcsin(x/2) so use this to change your limits of integration.

    Your final answer should be very nice (a multiple of pi in fact).
     
  4. Dec 12, 2008 #3
    Alternatively, interpret the integral geometrically and the answer is immediate.
     
  5. Dec 12, 2008 #4
    Hi,

    Thank you for the tips.

    [tex]x = 2 \ sin\theta[/tex]

    for [tex]x=2[/tex]:

    [tex]2 = 2 \ sin\theta[/tex]

    [tex]sin\theta=\frac{2}{2}=1[/tex]

    [tex]\theta=sin^{-1}(1)=\frac{\pi}{2}[/tex]

    for [tex]x=-2[/tex]

    [tex]-2=2 \ sin\theta[/tex]

    [tex]sin\theta = \frac{-2}{2}=-1[/tex]

    [tex]\theta=sin^{-1}(-1)=-\frac{\pi}{2}[/tex]

    Thus

    [tex]\int^{2}_{-2}\sqrt{4-x^2} \ dx = 4\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}cos^2\theta \ d\theta[/tex]

    [tex]\vdots[/tex]

    [tex]=2\theta+sin2\theta]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}[/tex]

    [tex]=[2(\frac{\pi}{2})+sin2(\frac{\pi}{2})]-[2(-\frac{\pi}{2})+sin2(-\frac{\pi}{2})][/tex]

    [tex]=(\pi+sin\pi})-[-\pi+sin(-\pi)][/tex]

    [tex]=\pi+sin\pi+\pi-sin(-\pi)[/tex]

    [tex]=\pi+sin\pi+\pi-(-sin\pi)[/tex]

    [tex]=\pi+sin\pi+\pi+sin\pi[/tex]

    [tex]=2\pi+2 \ sin\pi[/tex]

    [tex]=2\pi+2 \ (0)[/tex]

    [tex]=2\pi[/tex]

    [tex]=2 \ (180)[/tex]

    [tex]=360[/tex]

    Would anybody please evaluate it again

    Thanks
     
  6. Dec 12, 2008 #5
    If you convert radians to degrees, make it clear because [tex]\pi[/tex] does not equal 180.

    Use Anthony's advice to check your answer. Hint: the integral represents a semicircle.
     
  7. Dec 12, 2008 #6
    hmm...

    [tex]2\pi = 2\times3.14=6.28[/tex]

    Is that correct?

    Thanks for your help
     
  8. Dec 15, 2008 #7
    divided by 2 for half the area, so pi
     
    Last edited: Dec 15, 2008
  9. Dec 18, 2008 #8
    no lubuntu, he didn't need too, if you look at that integral, it's the area of a semicircle, if he had multiplied by 2 to get a full circle, he would have had to divide by 2, but he didn't, It looks correct as far as I can tell
     
  10. Dec 18, 2008 #9

    HallsofIvy

    User Avatar
    Science Advisor

    [tex]2\pi[/tex] is correct. The other two are only correct to two decimal places.
     
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