calculus integration

Question

Find or evaluate the integral by completing the square. (Round your answer to three decimal places.)
Boundary is from 2 to 3.

∫ (2x-3)/ (√(4x-x2) dx

Answer 1

Hi Nino

 

By completing the square you will find that 4x - x² can be written as 4 - (x-2)².

Also, the numerator can be written as 2(x-2) + 1.

 

Therefore your integral can be written as

 

∫ [2(x-2)+1] / √[4 - (x-2)²] dx

 

This calls for a trigonometric substitution using sin (since the expression in the denominator is of the form a² - x²). Specifically,

 

2sinθ = x - 2

 

Differentiating,

 

2cosθdθ = dx

 

Substituting into the integral:

 

∫ ([2(2sinθ) + 1] / √[4 - 4sin²θ]) 2cosθdθ  and your limits are now from 0 to π/6.

 

This simplifies to ∫(4sinθ + 1)dθ. (The square root simplifies to 2cosθ, which cancels out.)

 

Taking the antiderivative, you get -4cosθ + θ.

 

Evaluating from 0 to π/6 you get -2√3 + π/6 + 4, or about 1.059.

Answer 2

After completing the square, we have 

                       ∫_{(from 2 to 3)} [(2x-3)/√(4-(x-2)²)]dx

 

Use the substitution u = x-2 to obtain ∫_{(from 0 to 1)}[(2u+1)/√(4-u²)]du

 

Split the integral into:

 

     ∫_{(from 0 to 1)}[2u/√(4-u²)]du + ∫_{(from 0 to 1)}[1/√(4-u²)]du

 

     For the first integral, let w = 4-u².

 

       We get -∫_{(from 4 to 3)} w^-1/2dw = ∫_{(from 3 to 4)} w^-1/2dw

 

                                                = 2√w_{(from 3 to 4)} = 4 - 2√3

 

     So far we have 4-2√3 + ∫_{(from 0 to 1)}[1/√(4-u²)]du

 

                                   NOTE: ∫[1/√(a²-u²)]du = Sin^-1(u/a) + c

 

      So, we obtain 4 - 2√3 + Sin^-1(u/2)_{(from 0 to 1)}

 

                        = 4 - 2√3 + Sin^-1(1/2) - Sin^-1(0)

 

                        = 4 - 2√3 + π/6 - 0

 

                        = 4 - 2√3 + π/6

  

Answer 3

The "completing the square" part is done with the quadratic in the denominator:

   4x - x² = 0

   x² - 4x = 0

   x² - 4x + 4 = 4

   (x - 2)² = 4

   (x - 2)² - 4 = 0

 

Making this substitution into the integral, we get:

   ∫(2x - 3)/√[(x - 2)² - 4]

 

Now we want to use u-substitution with u = x - 2:

   ∫(2u + 1)/√(u² - 4)

 

Here is where the problem gets ugly. You have to split the integral into two parts and do two separate integrals:

   ∫2u/√(u² - 4) + ∫1/√(u² - 4)

 

The first integral is a simple substitution problem. The second one is a more difficult trig substitution. I will do the second integral below and leave you to do the first:

   ∫1/√(u² - 4)

   ∫1/√((2secθ)² - 4)

   ∫1/√(4sec²θ - 4)

   (1/2)*∫1/√(sec²θ - 1)

   (1/2)*∫1/√(tan²θ)

   (1/2)*∫1/tanθ

   (1/2)*∫cotθ

   (1/2)*ln|sinθ|

   (1/2)*ln|sin(sec^-1(u/2))|

   (1/2)*ln|sin(sec^-1((x-2)/2))|

 

From here you use the Fundamental Theorem of Calculus on the first integral (which you did yourself) PLUS this integral. I hope this helps.