Solve the indefinite integral of f(x)=(2x-5)/(x^2-5x+6) . Then solve the definite integral of f from x=0 to x=1.
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f(x) = (2x-5) /(x^2 -5x + 6)
F(x)= intg f(x) = intg (2x-5)/(x^2 - 5x + 6)
Let u = x^2 - 5x + 6
==> du = (2x - 5) dx
F(x) = intg du/u
= ln u
= ln...
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To evaluate the indefinit Int (2x-5)/(x^2-5x-6) dx .(ii) find the definite integral fro x= 0 to x=1.
Solution:
method (i) :
Pbviously numerator is diffrential coefficient of dinominator or
(x^2-5x-6)' = 2x-5.
So Int d(x^2-5x-6)/ (x^2-5x-6) = log(x^2-5x -6) + constant.
2nd method:
The denominator x^2-5x-6 = (x-6)(x+1)
So we split (2x-5)/(x^2-5x-6) into A/(x-6) +B/(x+1).
So (2x-5) = A(x+1) +B(x-6)....(1) Put x = 6, then
2*6-5 = A(6+1) + 0 . So 7 = 7A, So A = 1.
Put x = -1 in (1): 2*-1-5 = 0 + B(-1-6) . -7 = -7B. S B = a.
So 2x-5 = 1/(x-6) +1/(x+1)
Therefore Int (2x-5)/(x^2-5x-6) = Int dx/(x-6) +Int dx/(x-5)
Int (2x-5)/(x^2-5x-6) = log(x-6)+log(x+1) + Constant = log(x^2-5x-6) + C = F(x) say
To find the definite integral from x= 0 to x=1
F(1) - F(0) = log (1-5-6) - log (0-0-6) = ln(-10)/(-6) = ln(5/3) = 0.510825623.
To calculate the indefinite integral of the given function, we'll use the substitution method.
We'll calculate Integral of f(x) = (2x-5)/(x^2-5x+6)
We notice that if we'll differentiate the denominator, x^2-5x+6, we'll get 2x-5.
So, we'll note x^2-5x+6 = t
(x^2-5x+6)'dx = dt
(2x-5)dx = dt
We'll re-write the integral in the variable t:
Int (2x-5)dx/(x^2-5x+6) = Int dt / t
Int dt / t = ln t + C
But x^2-5x+6 = t.
Int (2x-5)dx/(x^2-5x+6) = ln(x^2-5x+6) + C, where C is a family of constants.
Now, we'll evaluate the definite integral, using Leibniz-Newton formula:
Int (2x-5)dx/(x^2-5x+6) = F(1) - F(0)
F(1) = ln(1^2-5*1+6)
F(1) = ln (1-5+6)
F(1) = ln 2
F(0) = ln(0^2-5*0+6)
F(0) = ln 6
Int (2x-5)dx/(x^2-5x+6) = ln 2 - ln 6
We'll apply the quotient rule to the difference of logarithms:
Int (2x-5)dx/(x^2-5x+6) = ln (2/6)
Int (2x-5)dx/(x^2-5x+6)=ln (1/3)=ln 1 - ln 3=0-ln 3=- ln 3
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