Cho hàm số \(f\left( x \right)\) có đạo hàm liên tục trên đoạn \(\left[ 1\,;2 \right]\) thỏa mãn \(\int\limits_{1}^{2}{{{\left( x-1 \right)}^{2}}f\left( x \right)\text{d}x}=-\frac{1}{3}\), \(f\left( 2 \right)=0\) và \(\int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}=7\). Tính tích phân \(I=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}\).

A. \(I=\frac{7}{5}\)

B. \(I=-\frac{7}{5}\).

C. \(I=-\frac{7}{20}\).

D. \(I=\frac{7}{20}\).

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Chọn B

Đặt

\(\left\{ \begin{align} & u=f\left( x \right) \\ & \text{d}v={{\left( x-1 \right)}^{2}}\text{d}x \\ \end{align} \right.\)\(\Rightarrow \left\{ \begin{align} & \text{d}u={f}'\left( x \right)\text{d}x \\ & v=\frac{{{\left( x-1 \right)}^{3}}}{3} \\ \end{align} \right.\)

Khi đó, \(\int\limits_{1}^{2}{{{\left( x-1 \right)}^{2}}f\left( x \right)\text{d}x}\)\( =\left. \frac{{{\left( x-1 \right)}^{3}}f\left( x \right)}{3} \right|_{1}^{2}-\)\( \frac{1}{3}\int\limits_{1}^{2}{{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}\)

\(\Rightarrow -\frac{1}{3}\)\( =-\frac{1}{3}\int\limits_{1}^{2}{{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}\)

\(\Rightarrow \int\limits_{1}^{2}{{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}=1\).

Ta lại có:

\(\left\{ \begin{align} & \int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}=7 \\ & \int\limits_{1}^{2}{14{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}=14 \\ & \int\limits_{1}^{2}{49{{\left( x-1 \right)}^{6}}\text{d}x}=\left. 7{{\left( x-1 \right)}^{7}} \right|_{1}^{2}=7 \\ \end{align} \right.\)

\(\Rightarrow \int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}\)\( -\int\limits_{1}^{2}{14{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}+\int\limits_{1}^{2}{49{{\left( x-1 \right)}^{6}}\text{d}x}=0\)

\(\Rightarrow \int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) -7{{\left( x-1 \right)}^{3}} \right]}^{2}}\text{d}x}=0\) \(\left( 1 \right)\),

mà \(\int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) -7{{\left( x-1 \right)}^{3}} \right]}^{2}}\text{d}x}\ge 0\).

nên \(\left( 1 \right)\Rightarrow {f}'\left( x \right) -7{{\left( x-1 \right)}^{3}}=0\)\( \Rightarrow {f}'\left( x \right)\)\( =7{{\left( x-1 \right)}^{3}}\)\( \Rightarrow f\left( x \right)\)\( =\frac{7{{\left( x-1 \right)}^{4}}}{4}+C\).

Mà \(f\left( 2 \right)=0\)\( \Leftrightarrow \frac{7}{4}+C=0\)\( \Leftrightarrow C=-\frac{7}{4}\)\( \Rightarrow f\left( x \right)\)\( =\frac{7}{4}\left[ {{\left( x-1 \right)}^{4}}-1 \right]\).

\(\Rightarrow I\)\( =\int\limits_{1}^{2}{f\left( x \right)\text{d}x}\)\( =\frac{7}{4}\int\limits_{1}^{2}{\left[ {{\left( x-1 \right)}^{4}}-1 \right]\text{d}x}\)\( =\left. \frac{7}{4}\left[ \frac{{{\left( x-1 \right)}^{5}}}{5}-x \right] \right|_{1}^{2}=-\frac{7}{5}\).

Vậy \(I=-\frac{7}{5}\).

Câu hỏi này thuộc đề thi trắc nghiệm dưới đây, bấm vào Bắt đầu thi để làm toàn bài