Tính tích phân \(\int\limits_0^1 {\max \left\{ {{e^x},{e^{1 – 2x}}} \right\}} dx\).

Tính tích phân \(\int\limits_0^1 {\max \left\{ {{e^x},{e^{1 – 2x}}} \right\}} dx\).

A. \(e – 1\).

B. \(\frac{3}{2}\left( {e – \sqrt[3]{e}} \right)\).

C. \(e – \sqrt[3]{e}\).

D. \(\frac{1}{2}\left( {e – \frac{1}{e}} \right)\).

Lời giải:

Ta có: \({e^x} \ge {e^{1 – 2x}} \Leftrightarrow x \ge 1 – 2x \Leftrightarrow x \ge \frac{1}{3}\). Suy ra: \(\max \left\{ {{e^x},{e^{1 – 2x}}} \right\} = \left\{ \begin{array}{l}{e^{1 – 2x}}\,\,khi\,\,0 \le x \le \frac{1}{3}\\{e^x}\,\,\,\,\,\,khi\,\,\frac{1}{3} \le x \le 1\,\end{array} \right.\)

Do đó \(I = \int\limits_0^1 {\max \left\{ {{e^x},{e^{1 – 2x}}} \right\}} dx = \int\limits_0^{\frac{1}{3}} {{e^{1 – 2x}}} dx + \int\limits_{\frac{1}{3}}^1 {{e^x}} dx = – \frac{1}{2}{e^{1 – 2x}}\left| {_{\scriptstyle\atop\scriptstyle0}^{\frac{1}{3}}} \right. + {e^x}\left| {_{\frac{1}{3}}^{\scriptstyle1\atop\scriptstyle}} \right.\)

\( = – \frac{1}{2}{e^{\frac{1}{3}}} + \frac{1}{2}e + e – {e^{\frac{1}{3}}} = \frac{3}{2}\left( {e – \sqrt[3]{e}} \right)\).

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Đây là các câu ÔN THI TN THPT MÔN TOÁN 2023 – CHUYÊN ĐỀ NGUYÊN HAM – TICH PHÂN – ỨNG DỤNG.