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?

$$\int_{-2}^{2} \sqrt{4 - x^2} \ dx$$

let

$$x = 2 sin \theta$$

then

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

also

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

thus

$$\sqrt{4-x^2}=\sqrt{4cos^2\theta}$$

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

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.