Cho hàm số f ( x ) = x ( x^2 + 3 ) . Xét I = 1 ∫ − 1 | f ( x ) | d x .

a) Đ, b) S, c) Đ, d) S

a) \({I_1} = \int\limits_0^1 {\left| {f\left( x \right)} \right|dx} \)\( = \int\limits_0^1 {\left| {x\left( {{x^2} + 3} \right)} \right|dx} \)\( = \int\limits_0^1 {x\left( {{x^2} + 3} \right)dx} \)\( = \int\limits_0^1 {\left( {{x^3} + 3x} \right)dx} \)\( = \left. {\left( {\frac{{{x^4}}}{4} + \frac{{3{x^2}}}{2}} \right)} \right|_0^1 = \frac{7}{4}\).

\({I_2} = \int\limits_{ - 1}^0 {\left| {f\left( x \right)} \right|dx} \)\( = \int\limits_{ - 1}^0 {\left| {x\left( {{x^2} + 3} \right)} \right|dx} \)\( = - \int\limits_{ - 1}^0 {x\left( {{x^2} + 3} \right)dx} \)\( = - \int\limits_{ - 1}^0 {\left( {{x^3} + 3x} \right)dx} \)\( = \left. { - \left( {\frac{{{x^4}}}{4} + \frac{{3{x^2}}}{2}} \right)} \right|_{ - 1}^0 = \frac{7}{4}\).

b) \(I = \int\limits_0^1 {\left| {f\left( x \right)} \right|dx} + \int\limits_{ - 1}^0 {\left| {f\left( x \right)} \right|dx} = \frac{7}{2}\).

c) \(J = \int\limits_0^m {\left| {f\left( x \right)} \right|dx} \)\( = \int\limits_0^m {\left| {x\left( {{x^2} + 3} \right)} \right|dx} \)\( = \int\limits_0^m {x\left( {{x^2} + 3} \right)dx} \)\( = \int\limits_0^m {\left( {{x^3} + 3x} \right)dx} \)

\( = \left. {\left( {\frac{{{x^4}}}{4} + \frac{{3{x^2}}}{2}} \right)} \right|_0^m = \frac{{{m^4}}}{4} + \frac{{3{m^2}}}{2}\).

Mà \(J = 4\) nên \(\frac{{{m^4}}}{4} + \frac{{3{m^2}}}{2} = 4\)\( \Leftrightarrow {m^4} + 6{m^2} - 16 = 0\)\( \Leftrightarrow m = \sqrt 2 \).

d) \(\int\limits_0^1 {x\left( {{x^2} + 3 - a\sqrt x } \right)dx = } \int\limits_0^1 {\left( {{x^3} + 3x - a{x^{\frac{3}{2}}}} \right)dx = } \left. {\left( {\frac{{{x^4}}}{4} + \frac{{3{x^2}}}{2} - \frac{2}{5}a{x^{\frac{5}{2}}}} \right)} \right|_0^1 = \frac{7}{4} - \frac{2}{5}a = 0\)\( \Leftrightarrow a = \frac{{35}}{8}\).