Cho hàm số f ( x ) = { x^2 − x khi x ≤ 0 x khi x > 0 . a) 5 ∫ 2 f ( x ) d x = 5 ∫ 2 x dx .

 a) \(\int\limits_2^5 {f\left( x \right)dx}  = \int\limits_2^5 {xdx} \).

b) \[\int\limits_{ - 4}^{ - 2} {f\left( x \right)dx}  = \int\limits_{ - 4}^{ - 2} {\left( {{x^2} - x} \right)dx}  = \left. {\left( {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right)} \right|_{ - 4}^{ - 2} = \frac{{74}}{3}\].

c) \(I = \int\limits_{ - 1}^1 {f\left( x \right)dx}  + \int\limits_{ - 1}^3 {f\left( x \right)dx}  = \int\limits_{ - 1}^0 {f\left( x \right)dx}  + \int\limits_0^1 {f\left( x \right)dx}  + \int\limits_{ - 1}^0 {f\left( x \right)dx}  + \int\limits_0^3 {f\left( x \right)dx} \)

\( = 2\int\limits_{ - 1}^0 {f\left( x \right)dx}  + \int\limits_0^1 {f\left( x \right)dx}  + \int\limits_0^3 {f\left( x \right)dx} \)

\( = 2\int\limits_{ - 1}^0 {\left( {{x^2} - x} \right)dx}  + \int\limits_0^1 {xdx}  + \int\limits_0^3 {xdx} \)

\( = \left. {2\left( {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right)} \right|_{ - 1}^0 + \left. {\frac{{{x^2}}}{2}} \right|_0^1 + \left. {\frac{{{x^2}}}{2}} \right|_0^3\)

\( = \frac{5}{3} + \frac{1}{2} + \frac{9}{2} = \frac{{20}}{3}\).

d) \(\int\limits_{ - 2}^1 {f\left( x \right)dx}  = \int\limits_{ - 2}^0 {f\left( x \right)dx}  + \int\limits_0^1 {f\left( x \right)dx} \)\( = \int\limits_{ - 2}^0 {\left( {{x^2} - x} \right)dx}  + \int\limits_0^1 {xdx} \)\( = \left. {\left( {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right)} \right|_{ - 2}^0 + \left. {\frac{{{x^2}}}{2}} \right|_0^1 = \frac{{14}}{3} + \frac{1}{2} = \frac{{31}}{6}\).

 Đáp án: a) Đúng;   b) Sai;   c) Đúng;   d) Đúng.