If for all real triplets $(a, b, c), f(x)=a+b x+c x^{2}$ then $\int \limits_{0}^{1} f(\mathrm{x}) \mathrm{d} \mathrm{x}$ is equal to
$\frac{1}{2}\left\{f(1)+3 f\left(\frac{1}{2}\right)\right\}$
$2\left\{3 f(1)+2 f\left(\frac{1}{2}\right)\right\}$
$\frac{1}{6}\left\{f(0)+f(1)+4 f\left(\frac{1}{2}\right)\right\}$
$\frac{1}{3}\left\{f(0)+f\left(\frac{1}{2}\right)\right\}$
Let $a, b, c$ be non-zero real numbers such that ; $\int\limits_0^1 {} (1 + cos^8x) (ax^2 + bx + c) dx$ $= \int\limits_0^2 {} (1 + cos^8x) (ax^2 + bx + c) dx$ , then the quadratic equation $ax^2 + bx + c = 0$ has :
$\int\limits_{ - 1}^{\frac{3}{2}} {|x\sin \pi x|dx} $ equals
The numbers $P, Q$ and $R$ for which the function $f(x) = P{e^{2x}} + Q{e^x} + Rx$ satisfies the conditions $f(0) = - 1,$ $f'(\log 2) = 31$ and $\int_0^{\log 4} {[f(x) - Rx]\,dx = \frac{{39}}{2}} $ are given by
The value of integral $\int_0^1 {\frac{{{x^b} - 1}}{{\log x}}} \,dx$ is
Let $f$ be a positive function. Let
${I_1} = \int_{1 - k}^k {x\,f\left\{ {x(1 - x)} \right\}} \,dx$, ${I_2} = \int_{1 - k}^k {\,f\left\{ {x(1 - x)} \right\}} \,dx$
when $2k - 1 > 0.$ Then ${I_1}/{I_2}$ is