Câu hỏi

17/12/2024 1

Tính các tích phân sau:

a) \(\int\limits_0^2 {\left| {2x – 1} \right|dx} \);

b) \(\int\limits_{ – 2}^3 {\left| {x – 1} \right|dx} \).

Danh mục liên quan

  • Trắc Nghiệm Toán 12
  • Lời giải của Vua Trắc Nghiệm

    a) Ta có \(\int\limits_0^2 {\left| {2x – 1} \right|dx} \) = \(\int\limits_0^{\frac{1}{2}} {\left| {2x – 1} \right|dx} + \int\limits_{\frac{1}{2}}^2 {\left| {2x – 1} \right|dx} \)

    = \(\int\limits_0^{\frac{1}{2}} {\left( {1 – 2x} \right)dx} + \int\limits_{\frac{1}{2}}^2 {\left( {2x – 1} \right)dx} \)

    = \(\left. {\left( {x – {x^2}} \right)} \right|_0^{\frac{1}{2}} + \left. {\left( {{x^2} – x} \right)} \right|_{\frac{1}{2}}^2\) = \(\frac{5}{2}\).

    b) Ta có \(\int\limits_{ – 2}^3 {\left| {x – 1} \right|dx} \) = \(\int\limits_{ – 2}^1 {\left| {x – 1} \right|dx} + \int\limits_1^3 {\left| {x – 1} \right|dx} \)

    = \(\int\limits_{ – 2}^1 {\left( {1 – x} \right)dx} + \int\limits_1^3 {\left( {x – 1} \right)dx} \)

    = \(\left. {\left( {x – \frac{1}{2}{x^2}} \right)} \right|_{ – 2}^1 + \left. {\left( {\frac{1}{2}{x^2} – x} \right)} \right|_1^3\)

    = 12 − \(\frac{1}{2}\).12 – (−2) + \(\frac{1}{2}\).(−2)2 + \(\frac{1}{2}\).32 – 3 − \(\frac{1}{2}\).12 + 1.

    = \(\frac{{13}}{2}\).

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