Cho hàm số \(f(x) = \left\{ \begin{array}{l}2{x^2} – 1\,\,\,\,\,\,{\rm{khi}}\,\,x < 0\\x – 1\,\,\,\,\,\,\,\,\,\,\,{\rm{khi}}\,\,0 \le x \le 2\\5 – 2x\,\,\,\,\,\,\,\,{\rm{khi}}\,\,x > 2\,\end{array} \right.\). Tính tích phân \(\int\limits_{ – \frac{\pi }{4}}^{\frac{\pi }{4}} {f\left( {2 – 7\tan x} \right)} \frac{1}{{{{\cos }^2}x}}{\rm{d}}x\).

DẠNG TOÁN 41 TÍCH PHÂN HÀM ẨN – phát triển theo đề tham khảo Toán 2021

 

Theo đề tham khảo Toán 2021 của Bộ GD&ĐT

ĐỀ BÀI:

Cho hàm số \(f(x) = \left\{ \begin{array}{l}2{x^2} – 1\,\,\,\,\,\,{\rm{khi}}\,\,x < 0\\x – 1\,\,\,\,\,\,\,\,\,\,\,{\rm{khi}}\,\,0 \le x \le 2\\5 – 2x\,\,\,\,\,\,\,\,{\rm{khi}}\,\,x > 2\,\end{array} \right.\). Tính tích phân \(\int\limits_{ – \frac{\pi }{4}}^{\frac{\pi }{4}} {f\left( {2 – 7\tan x} \right)} \frac{1}{{{{\cos }^2}x}}{\rm{d}}x\).

A.\(\frac{{201}}{{77}}\).

B. \(\frac{{34}}{{103}}\).

C. \(\frac{{155}}{7}\).

D. \(\frac{{109}}{{21}}\).

GY::

Xét \(I = \int\limits_{ – \frac{\pi }{4}}^{\frac{\pi }{4}} {f\left( {2 – 7\tan x} \right)\frac{1}{{{{\cos }^2}x}}{\rm{d}}x} \)

Đặt \(2 – 7\tan x = t \Rightarrow \frac{1}{{{{\cos }^2}x}}{\rm{d}}x =  – \frac{1}{7}{\rm{d}}t\)

Với \(x =  – \frac{\pi }{4}\)\( \Rightarrow \)\(t = 9\)

\(x = \frac{\pi }{4}\)\( \Rightarrow \)\(t =  – 5\)

\( \Rightarrow I = \frac{1}{7}\int\limits_{ – 5}^9 {f\left( t \right){\rm{d}}t}  = \frac{1}{7}\int\limits_{ – 5}^9 {f\left( x \right){\rm{d}}x}  = \frac{1}{7}\int\limits_{ – 5}^0 {f(x){\rm{d}}x}  + \frac{1}{7}\int\limits_0^2 {f(x){\rm{d}}x}  + \frac{1}{7}\int\limits_2^9 {f(x){\rm{d}}x} \)

\( = \frac{1}{7}\int\limits_{ – 5}^0 {\left( {2{x^2} – 1} \right){\rm{d}}x}  + \frac{1}{7}\int\limits_0^2 {\left( {x – 1} \right){\rm{d}}x}  + \frac{1}{7}\int\limits_2^9 {\left( {5 – 2x} \right){\rm{d}}x}  = \frac{{109}}{{21}}.\)