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how do u deal with the 1 in integral(from 0 to 3) (1+sqrt(9-x^2)dx

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4 Answers

Hi Rhonda!

To deal with the 1 in this integral, you can break up the original integral as follows:

∫[1+√(9-x^2)]dx = ∫1dx + ∫√(9-x^2)dx

*Note: I have indicated these as indefinite integrals because there is no way to nicely format definite integrals on here.

From there, you take the definite integrals each separately as you normally would, and add them together to find the answer.

Hope this helps! Let me know if you have any further questions.

-Leigh

Comments

Good answer!

It may not be apparent to the OP, but ?v(9-x2)dx isn't a simple one, as you can't use u-substitution.  There's two ways to handle this.  First is to realize that this is the equation of a circle with radius=3, and thus the integral is asking for the area in the first quadrant.  So you can just calculate area of the circle and divide it by 4 - that's the "graphical" method.

To actually integrate this, the simplest method is to convert it to polar form:

y = v(9-x2)
y2 = 9 - x2
x2 + y2 = 9
r2 = 9
r = 3
? (1/2)(3)2 d?   (from 0 -> pi/2)

Wow this editor doesn't work well at all...    It messed up a bunch of characters.  That last line should read "integral of (1/2)32 times d-theta"  

Test: ? θ θ T

 

Rhonda,

This problem can be done much simpler if you notice that the second part of the integral is in the first quadrant and is equal to a quarter of circle's area with r = 3. So,

∫(from 0 to 3) (1+sqrt(9-x^2)dx = 3 + (1/4)pi(3^2)

Remember the rule ∫ ( f(x) + g(x) )dx = ∫ f(x)dx + ∫ g(x)dx. In other words, integration is additive (distributes over addition). That is how the 1 is pulled out (see Leigh's answer.)

If you really wanted to use symbolic integration and plug in the bounds then use trigonometric substitution. Some of you tried to show this but had formatting trouble.

If x = 3 sin θ then dx = 3 cos θ dθ

We get

   ∫ [1+√(9-x2)] dx

= ∫ [1+√(9-9 sin2 θ)] * 3 cos θ dθ

= ∫ [3 cos θ + 9 cos2 θ] dθ

= ∫ [3 cos θ + 9(1+cos 2θ)/2 ] dθ

= ∫ [3 cos θ + 9/2 + (9/2) cos 2θ] dθ

= 3 sin θ + 9θ/2 + (9/4) sin 2θ + C

= 3 sin θ + 9θ/2 + (9/2) sin θ cos θ+ C

= x + (9/2) sin-1 (x/3) + (9/2) (x/3) √(1 - x2/9) + C

= x + (9/2) sin-1 (x/3) + (x/2)√(9 - x2) + C

Substituting the bounds -3 and 3 gives

[3+(9/2) sin-1 (3/3)+(1/2) (3) √(9 - 32)] - [0+(9/2) sin-1 (0/3)+(1/2) (0) √(9 - 02)]

= [3+(9/2) (π/2) + 0] - [0 + (9/2) (0) + 0]

= 3+9π/4

Same as the answer realized with the recognition that the integral equals the sum of a rectangle and quarter-circle.

Comments

Roman, not quite correct....   Your main error is a simple one - the limits are 0..3, not -3..3.  The answer should only be 3+9pi/4.

Second, to be clear, the original equation is NOT a semi-circle, it is a circlular-curve shifted up by 1.  That is, if you graph sqrt(9-x2) then you don't get a circle, you only get the positive half of the circle (above the x-axis).  This graph is then shifted up by 1.  And then, you only consider from 0..3.  The area we are considering is a "quarter-circle or radius 3 sitting on top of a 3x1 rectangle in the first quadrant".

Otherwise, your math is correct, although somewhat over-complicated and prone to mistakes.  A cleaner method is to convert to polar coordinates, as I said in my other comment:

A = 3 + Integral from 0..3 of 1/2 32 dTheta  

(apologies for the use of words - I've discovered that the symbols aren't rendered correctly in comments.)  The '3' comes from the area of the rectangle as a result of splitting the integral into f(x)=1 and g(x)=sqrt(9-x2) and the Integral from 0..3 1/2(32)dTheta is the area of the unshifted circle in the first quadrant.  This gives 3+9pi/4, the correct answer.

 

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