Determine the value of a (a>0) if:

Question

∫⁷_a (-1/1-7x)dx = (3/7)ln3

Answer 1

Since this is a definite integral we can apply the rules of integration and affect the boundary conditions to determine the equation for a which must solve for (3/8)ln(3)...

int_[lb=a]_[ub=7] (-1)/(1-7x) dx can be transformed into a simple integral using u-substitution

Let u = 7x - 1 = -1(1-7x)

This implies du = 7dx or dx = du/7 --> [ du/dx = 7 ]

Now let's rewrite the boundaries of the integral in terms of u instead of x.

when x = 7 --> u = 48

when x = a --> u = 7a-1

int_[lb=7a-1]_[ub=48] (1/7)*(1/u)*du --> 1/7ln(u)_lb=7a-1]_[ub=48]

---> (1/7)ln(48) - (1/7)ln(7a-1)

Due to the properties of logarithms --> ln(a) - ln(b) = ln(a/b)

(1/7)ln(48) - (1/7)ln(7a-1) = (1/7)ln(48/(7a-1))

This must match the solution provided which is (3/7)ln(3)

(1/7)ln(48/(7a-1)) = (3/7)ln(3) --> ln(48/(7a-1)) = 3ln(3)

3ln(3) --> ln(3³) = ln(27)

ln(48/(7a-1)) - ln(27) --> 48/(7a-1) = 27

48/27 = 7a -1

16/9 = 7a - 1

25/9 = 7a

Therefore the solution for a is...

a = 25/63