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, $$\int\limits_{}^{} {f(x)dx} $$
$$ = \int\limits_{}^{} {(1 + 2x + 3{x^2} + 4{x^3})dx} $$
$$ = x + {x^2} + {x^3} + {x^4}$$
$$ \Rightarrow \int\limits_0^{1/2} {f(x)dx = {{15} \over {16}} > {3 \over 4}} $$
$$\int\limits_0^{3/4} {f(x)dx = {{530} \over {256}} < 3} $$
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 real numbers lies in the interval
Given, $$f(x) = 4{x^3} + 3{x^2} + 2x + 1$$
$$f'(x) = 2(6{x^2} + 3x + 1)$$
$$D = 9 - 24 < 0$$
Hence, f(x) = 0 has only one real root.
$$f\left( { - {1 \over 2}} \right) = 1 - 1 + {3 \over 4} - {4 \over 8} > 0$$
$$f\left( { - {3 \over 4}} \right) = 1 - {6 \over 4} + {{27} \over {16}} - {{108} \over {64}}$$
$$ = {{64 - 96 + 108 - 108} \over {64}} < 0$$
f(x) changes its sign in $$\left( { - {3 \over 4},{{ - 1} \over 2}} \right)$$,
hence f(x) = 0 has a root in $$\left( { - {3 \over 4},{{ - 1} \over 2}} \right)$$.
$$\int\limits_0^1 {{{{x^4}{{(1 - x)}^4}} \over {1 + {x^2}}}dx = \int\limits_0^1 {{{{x^4}{{\{ (1 + {x^2}) - 2x\} }^2}} \over {1 + {x^2}}}dx} } $$
$$ = \int\limits_0^1 {{x^4}{{{{(1 + {x^2})}^2} - 4x(1 + {x^2}) + 4{x^2}} \over {1 + {x^2}}}dx} $$
$$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + {{4\{ 1 + {x^2} - 1\} } \over {1 + {x^2}}}} \right]dx} $$
$$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + 4 - {4 \over {1 + {x^2}}}} \right]dx} $$
$$ = \int\limits_0^1 {\left[ {{x^6} - 4{x^5} + 5{x^4} - 4{{{x^4} - 1 + 1} \over {1 + {x^2}}}} \right]dx} $$
$$ = \int\limits_0^1 {({x^6} - 4{x^5} + 5{x^4} - 4{x^2} + 4)dx - 4\int\limits_0^1 {{{dx} \over {1 + {x^2}}}} } $$
$$ = \left[ {{{{x^7}} \over 7} - {{2{x^6}} \over 3} + {x^5} - {{4{x^3}} \over 3} + 4x} \right]_0^1 - 4[{\tan ^{ - 1}}x]_0^1$$
$$ = \left[ {{1 \over 7} - {2 \over 3} + 1 - {4 \over 3} + 4} \right] - \pi = {{22} \over 7} - \pi $$
$$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^3}}}\int_0^x {{{t\log (1 + t)} \over {4 + {t^4}}}dt} $$
Using L' Hospital's rule,
$$ = \mathop {\lim }\limits_{x \to 0} {{{{x\log (1 + x)} \over {4 + {x^4}}}} \over {3{x^2}}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + x)} \over {3x}}\,.\,{1 \over {4 + {x^4}}}$$
$$ = {1 \over 3}\,.\,{1 \over 4} = {1 \over {12}}$$ [using, $$\mathop {\lim }\limits_{x \to 0} {{\log (1 + x)} \over x} = 1$$]
