Tính: a) tích phân từ -1 đến 1/2 của (4x^3 -5) dx - tích phân từ 1 đến 1/2 của (4x^3 -5) dx

a) \(\int\limits_{ - 1}^{\frac{1}{2}} {\left( {4{x^3} - 5} \right)dx} - \int\limits_1^{\frac{1}{2}} {\left( {4{x^3} - 5} \right)dx} \)\( = \int\limits_{ - 1}^{\frac{1}{2}} {\left( {4{x^3} - 5} \right)dx} + \int\limits_{\frac{1}{2}}^1 {\left( {4{x^3} - 5} \right)dx} \)\( = \int\limits_{ - 1}^1 {\left( {4{x^3} - 5} \right)dx} \)

\( = \left. {\left( {{x^4} - 5x} \right)} \right|_{ - 1}^1\)\( = - 4 - 6 = - 10\).

b) \(\int\limits_0^3 {\left| {x - 1} \right|dx} \)\( = \int\limits_0^1 {\left| {x - 1} \right|dx} + \int\limits_1^3 {\left| {x - 1} \right|dx} \)\( = \int\limits_0^1 {\left( {1 - x} \right)dx} + \int\limits_1^3 {\left( {x - 1} \right)dx} \)

\( = \left. {\left( {x - \frac{{{x^2}}}{2}} \right)} \right|_0^1 + \left. {\left( {\frac{{{x^2}}}{2} - x} \right)} \right|_1^3\)\( = \frac{1}{2} + \frac{3}{2} + \frac{1}{2} = \frac{5}{2}\).

c) \(\int\limits_0^\pi {\left| {\cos x} \right|dx} \)\( = \int\limits_0^{\frac{\pi }{2}} {\left| {\cos x} \right|dx} + \int\limits_{\frac{\pi }{2}}^\pi {\left| {\cos x} \right|dx} \)\( = \int\limits_0^{\frac{\pi }{2}} {\cos xdx} - \int\limits_{\frac{\pi }{2}}^\pi {\cos xdx} \)

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