Bài 2: Tích phân

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datcoder

Tính:

a) \(\int\limits^{\dfrac{1}{2}}_{-1}\left(4x^3-5\right)dx-\int\limits^{\dfrac{1}{2}}_1\left(4x^3-5\right)dx;\)                 b) \(\int\limits^3_0\left|x-1\right|dx;\)              c) \(\int\limits^{\pi}_0\left|cosx\right|dx.\)

datcoder
29 tháng 10 lúc 22:59

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

\( = 4\int\limits_{ - 1}^1 {{x^3}dx}  - 5\int\limits_{ - 1}^1 {dx}  = \left. {\left( {{x^4}} \right)} \right|_{ - 1}^1 - 5\left. {\left( x \right)} \right|_{ - 1}^1 = \left[ {{1^4} - {{\left( { - 1} \right)}^4}} \right] - 5\left[ {1 - \left( { - 1} \right)} \right] =  - 10\)

b) \(\int\limits_0^3 {\left| {x - 1} \right|dx}  = \int\limits_0^1 {\left| {x - 1} \right|dx}  + \int\limits_1^3 {\left| {x - 1} \right|dx}  = \int\limits_0^1 {\left( {1 - x} \right)dx}  + \int\limits_1^3 {\left( {x - 1} \right)dx}  = \left. {\left( {x - \frac{{{x^2}}}{2}} \right)} \right|_0^1 + \left. {\left( {\frac{{{x^2}}}{2} - x} \right)} \right|_1^3\)

\( = \left[ {\left( {1 - \frac{{{1^2}}}{2}} \right) - \left( {0 - \frac{{{0^2}}}{2}} \right)} \right] + \left[ {\left( {\frac{{{3^2}}}{2} - 3} \right) - \left( {\frac{{{1^2}}}{2} - 1} \right)} \right] = \frac{1}{2} + 2 = \frac{5}{2}\)

c)  \(\int\limits_0^\pi  {\left| {\cos x} \right|dx}  = \int\limits_0^{\frac{\pi }{2}} {\cos xdx}  + \int\limits_{\frac{\pi }{2}}^\pi  {\left( { - \cos x} \right)dx}  = \int\limits_0^{\frac{\pi }{2}} {\cos xdx}  - \int\limits_{\frac{\pi }{2}}^\pi  {\cos xdx}  = \left. {\left( {\sin x} \right)} \right|_0^{\frac{\pi }{2}} - \left. {\left( {\sin x} \right)} \right|_{\frac{\pi }{2}}^\pi \)

\( = \left( {\sin \frac{\pi }{2} - \sin 0} \right) - \left( {\sin \pi  - \sin \frac{\pi }{2}} \right) = 2\)