Cho số thực a và hàm số f ( x ) = { 2 x khi x ≤ 0 a ( x − x^2 ) khi x > 0 .
a) \(\int\limits_{ - 1}^0 {f\left( x \right)} {\rm{d}}x = \int\limits_{ - 1}^0 {2x} {\rm{d}}x\).
b) \[\int\limits_0^1 {f\left( x \right)} {\rm{d}}x = \int\limits_0^1 {a\left( {x - {x^2}} \right)} {\rm{d}}x\]\[ = \left. {a\left( {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = \frac{a}{6}\].
c) Với \(a = 2\) thì \(\int_{ - 1}^1 {f\left( x \right)dx} = \int_{ - 1}^0 {f\left( x \right)dx} + \int_0^1 {f\left( x \right)dx} = \left. {{x^2}} \right|_{ - 1}^0 + \frac{2}{6} = - 1 + \frac{2}{6} = \frac{{ - 2}}{3}\).
d) \(\int_{ - 1}^2 {f\left( x \right)dx} > 3\)\( \Leftrightarrow \int_{ - 1}^0 {f\left( x \right)dx} + \int_0^2 {f\left( x \right)dx} > 3\)\( \Leftrightarrow \int_{ - 1}^0 {2xdx} + \int_0^2 {a\left( {x - {x^2}} \right)dx} > 3\)
\( \Leftrightarrow \left. {{x^2}} \right|_{ - 1}^0 + \left. {a\left( {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3}} \right)} \right|_0^2 > 3\)\( \Leftrightarrow - 1 - \frac{{2a}}{3} > 3\)\( \Leftrightarrow a < - 6\).
Đáp án: a) Đúng; b) Sai; c) Đúng; d) Sai.