We can write the integral
$$\int\limits_0^b {{{(1 - x)}^2}dx - \int\limits_0^1 {{{(1 - x)}^2}dx = {1 \over 4}} } $$
$$ \Rightarrow \left. {{{{{(x - 1)}^3}} \over 3}} \right|_0^b - \left. {{{{{(x - 1)}^3}} \over 3}} \right|_b^1 = {1 \over 4}$$
$$ \Rightarrow {{{{(b - 1)}^3}} \over 3} + {1 \over 3} - \left( {0 - {{{{(b - 1)}^3}} \over 3}} \right) = {1 \over 4}$$
$$ \Rightarrow {{2{{(b - 1)}^3}} \over 3} = - {1 \over {12}} \Rightarrow {(b - 1)^3} = - {1 \over 8} \Rightarrow b = {1 \over 2}$$
$${x^2} = t \Rightarrow 2x\,dx = dt$$
$$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin t} \over {\sin t + \sin (\ln 6 - t)}}dt} $$ and $$I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {{{\sin (\ln 6 - t)} \over {\sin (\ln 6 - t) + \sin t}}dt} $$
$$2I = {1 \over 2}\int\limits_{\ln 2}^{\ln 3} {1dt \Rightarrow I = {1 \over 4}\ln {3 \over 2}} $$.
Consider the polynomial
$$f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}.$$
Let $$s$$ be the sum of all distinct real roots of $$f(x)$$ and let $$t = \left| s \right|.$$
The function$$f'(x)$$ is
Consider the polynomial
$$f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}.$$
Let $$s$$ be the sum of all distinct real roots of $$f(x)$$ and let $$t = \left| s \right|.$$
The area bounded by the curve $$y=f(x)$$ and the lines $$x=0,$$ $$y=0$$ and $$x=t,$$ lies in the interval
$$\int\limits_0^{1/2} {f(x)dx < \int\limits_0^t {f(x)dx < \int\limits_0^{3/4} {f(x)dx} } } $$
Now, $$\int\limits_{}^{} {f(x)dx} $$
$$ = \int\limits_{}^{} {(1 + 2x + 3{x^2} + 4{x^3})dx} $$
$$ = x + {x^2} + {x^3} + {x^4}$$
$$ \Rightarrow \int\limits_0^{1/2} {f(x)dx = {{15} \over {16}} > {3 \over 4}} $$
$$\int\limits_0^{3/4} {f(x)dx = {{530} \over {256}} < 3} $$
