Evaluate Integral: \int_{-2}^{2} \sqrt{4 - x^2} \ dx

[newabb]

Hi,

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

<br /> \int_{-2}^{2} \sqrt{4 - x^2} \ dx<br />

let

x = 2 sin \theta

then

\frac{dx}{d\theta}=2cos\theta, \ dx=2cos\theta \ d\theta

also

4-x^2 = 4-(2sin\theta)^2<br /> = 4 - 4sin^2\theta = 4(1-sin^2\theta)=4cos^2\theta<br />

thus

<br /> \sqrt{4-x^2}=\sqrt{4cos^2\theta}<br />

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

let u=2\theta then \frac{du}{d\theta}=2, \ d\theta = \frac{1}{2}du

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

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

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

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

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

= 4 + sin4 + 4 - sin(-4)

= 8 + sin4 - sin(-4)

= 8 + 0.07 - (-0.035)

= 8 + 0.07 + 0.035

= 8 + 0.07 + 0.035

= 8.105

Thenk you very much for evaluating

 

[snipez90]

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 2\int_{0}^{2} \sqrt{4 - x^2} \ dx 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).

 

[Anthony]

Alternatively, interpret the integral geometrically and the answer is immediate.

 

[newabb]

Hi,

Thank you for the tips.

x = 2 \ sin\theta

for x=2:

2 = 2 \ sin\theta

sin\theta=\frac{2}{2}=1

\theta=sin^{-1}(1)=\frac{\pi}{2}

for x=-2

-2=2 \ sin\theta

sin\theta = \frac{-2}{2}=-1

\theta=sin^{-1}(-1)=-\frac{\pi}{2}

Thus

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

\vdots

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

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

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

=\pi+sin\pi+\pi-sin(-\pi)

=\pi+sin\pi+\pi-(-sin\pi)

=\pi+sin\pi+\pi+sin\pi

=2\pi+2 \ sin\pi

=2\pi+2 \ (0)

=2\pi

=2 \ (180)

=360

Would anybody please evaluate it again

Thanks

 

[mutton]

If you convert radians to degrees, make it clear because \pi does not equal 180.

Use Anthony's advice to check your answer. Hint: the integral represents a semicircle.

 

[newabb]

hmm...

2\pi = 2\times3.14=6.28

Is that correct?

Thanks for your help

 

[lubuntu]

divided by 2 for half the area, so pi

 

[samdunhamss]

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

 

[HallsofIvy]

hmm...

2\pi = 2\times3.14=6.28

Is that correct?

Thanks for your help

2\pi is correct. The other two are only correct to two decimal places.