# What is the set A={(a,b)| a>1,b>1,a+b=7, I(a,b)=ln(b-1)/(a-1)} ?I(a,b) is the antiderivative of the function 1/square root(1+x^2), x is in the interval [a/b,b/a].

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### 1 Answer

First, we'll deal with the definite integral of the function f(x) = 1/sqrt(1 + x^2)

Int f(x)dx = Int dx/sqrt(1 + x^2)

According to the formula, the primitive of the function 1//sqrt(1 + x^2) is F(x) = ln[x + sqrt(1+x^2)]

Int dx/sqrt(1 + x^2) = ln[x + sqrt(1+x^2)]

But, according to Leibniz-Newton, the definite integral of the function is:

Int f(x)dx = F(b/a) - F(a/b)

F(b/a) - F(a/b) = ln[b + sqrt(a^2+b^2)]/a - ln[a + sqrt(a^2+b^2)]/b

We'll apply the quotient rule:

ln[b + sqrt(a^2+b^2)]/a - ln[a + sqrt(a^2+b^2)]/b = ln{ b[b + sqrt(a^2+b^2)]/a [a + sqrt(a^2+b^2)]} (1)

From the constraint imposed by enunciation, we'll get:

I(a,b) = ln(b-1)/(a-1) (2)

We'll equate (1) and (2):

ln{ b[b + sqrt(a^2+b^2)]/a [a + sqrt(a^2+b^2)]} = ln(b-1)/(a-1)

Since the bases are matching, we'll apply one to one rule:

b[b + sqrt(a^2+b^2)]/a [a + sqrt(a^2+b^2)] = (b-1)/(a-1)

We'll remove the brackets and cross multiply:

ab^2 + ab*sqrt(a^2+b^2) - b^2 - b*sqrt(a^2+b^2)] = ba^2 + ab*sqrt(a^2+b^2) - a^2 - a*sqrt(a^2+b^2)]

ab(b-a) - (b-a)(b+a) - (b-a)sqrt(a^2+b^2) = 0

Comparing, we'll get:

b - a = 0

ab - (a+b) - sqrt(a^2+b^2) = 0

We'll get the systems of equations:

a + b = 7 (1)

b - a = 0 (2)

Adding (1) and (2):

2b = 7 => b = 7/2 = b

We'll solve the next system of equation:

a + b = 7 (3)

ab - (a+b) - sqrt(a^2+b^2) = 0 (4)

ab - 7 - sqrt(49 - 2ab) = 0 =>ab - 7 = sqrt(49 - 2ab)

We'll raise to square:

(ab - 7)^2 = 49 - 2ab

We'll exand the binomial:

(ab)^2 - 14ab + 49 = 49 - 2ab

(ab)^2- 12ab = 0

ab(ab - 12) = 0

Finally, we'll get:

a + b = 7 and ab = 12, since ab cannot be zero, because of the constraints of enunciation that a>1 and b>1.

We'll get the quadratic:

x^2 - (a+b)x + ab = 0

x^2 - 7x + 12 = 0

a = 4 and b = 3 or a = 3 and b = 4.

**The set A is formed from the following pairs: {(7/2 , 7/2) ; (4 , 3) ; (3 , 4)}.**