Đề bài

Tính các tích phân sau: a) \(\int\limits_{ - 2}^1 {\left| {2x + 2} \right|dx} \) b) \(\int\limits_0^4 {\left| {{x^2} - 4} \right|dx} \) c) \(\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\sin x} \right|dx} \)

Lời giải chi tiết

a) \(\int\limits_{ - 2}^1 {\left| {2x + 2} \right|dx}  = \int\limits_{ - 2}^{ - 1} {\left| {2x + 2} \right|dx}  + \int\limits_{ - 1}^1 {\left| {2x + 2} \right|dx}  = \int\limits_{ - 2}^{ - 1} { - \left( {2x + 2} \right)dx}  + \int\limits_{ - 1}^1 {\left( {2x + 2} \right)dx} \) \( =  - \left. {\left( {{x^2} + 2x} \right)} \right|_{ - 2}^{ - 1} + \left. {\left( {{x^2} + 2x} \right)} \right|_{ - 1}^1 =  - \left[ {\left( { - 1} \right) - 0} \right] + \left[ {3 - \left( { - 1} \right)} \right] = 5\) b) \(\int\limits_0^4 {\left| {{x^2} - 4} \right|dx}  = \int\limits_0^2 {\left| {{x^2} - 4} \right|dx}  + \int\limits_2^4 {\left| {{x^2} - 4} \right|dx}  = \int\limits_0^2 {\left( {4 - {x^2}} \right)dx}  + \int\limits_2^4 {\left( {{x^2} - 4} \right)dx} \) \( = \left. {\left( {4x - \frac{{x{\rm{\^3}}}}{3}} \right)} \right|_0^2 + \left. {\left( {\frac{{x{\rm{\^3}}}}{3} - 4x} \right)} \right|_2^4 = \left( {\frac{{16}}{3} - 0} \right) + \left[ {\frac{{16}}{3} - \left( { - \frac{{16}}{3}} \right)} \right] = 16\) c) \(\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\sin x} \right|dx}  = \int\limits_{ - \frac{\pi }{2}}^0 {\left| {\sin x} \right|dx}  + \int\limits_0^{\frac{\pi }{2}} {\left| {\sin x} \right|dx}  = \int\limits_{ - \frac{\pi }{2}}^0 {\left( { - \sin x} \right)dx}  + \int\limits_0^{\frac{\pi }{2}} {\sin xdx} \) \( =  - \left. {\left( { - \cos x} \right)} \right|_{ - \frac{\pi }{2}}^0 + \left. {\left( { - \cos x} \right)} \right|_0^{\frac{\pi }{2}} =  - \left[ { - 1 - 0} \right] + \left[ {0 - \left( { - 1} \right)} \right] = 2\)