Đề cương ôn tập giữa kì 2 Toán 12 Chân trời sáng tạo cấu trúc mới (có tự luận) có đáp án - Tự luận

a) Ta có \(\int {f\left( x \right)} dx = \int {\left( {\frac{1}{3}{x^3} - 2{x^2} + x - 2024} \right)dx} \)\( = \frac{{{x^4}}}{{12}} - \frac{2}{3}{x^3} + \frac{{{x^2}}}{2} - 2024x + C\).

b) \(f\left( x \right) = \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) = {x^3} + 6{x^2} + 11x + 6\).

Suy ra \(\int {f\left( x \right)} dx = \int {\left( {{x^3} + 6{x^2} + 11x + 6} \right)dx} = \frac{{{x^4}}}{4} + 2{x^3} + \frac{{11}}{2}{x^2} + 6x + C\).

c) Ta có \(f\left( x \right) = {\left( {x - 1} \right)^2} = {x^2} - 2x + 1\).

Suy ra \(\int {f\left( x \right)dx} = \int {\left( {{x^2} - 2x + 1} \right)dx} = \frac{{{x^3}}}{3} - {x^2} + x + C\).

a) \(\int {f\left( x \right)dx} = \int {\left( {2\sin x + 3x} \right)dx} = - 2\cos x + \frac{3}{2}{x^2} + C\).

b) \[\int {f\left( x \right)dx} = \int {{{\cos }^2}\frac{x}{2}dx} = \int {\frac{{1 + \cos x}}{2}dx} = \frac{1}{2}\int {\left( {1 + \cos x} \right)dx} = \frac{1}{2}\left( {x + \sin x} \right) + C\].

c) \(\int {f\left( x \right)dx} = \int {\sin 3xdx} = \frac{1}{3}\int {\sin 3xd\left( {3x} \right)} = - \frac{{\cos 3x}}{3} + C\).

a) \(\int {f\left( x \right)dx} = \int {\frac{{{x^2} + 3x + 2}}{x}dx} = \int {\left( {x + 3 + \frac{2}{x}} \right)dx} = \frac{{{x^2}}}{2} + 3x + 2\ln \left| x \right| + C\).

b) \(\int {f\left( x \right)dx} = \int {\frac{{{x^2} + 3x + 1}}{{\sqrt x }}} dx = \int {\left( {x\sqrt x + 3\sqrt x + \frac{1}{{\sqrt x }}} \right)dx} \)

\( = \int {\left( {{x^{\frac{3}{2}}} + 3{x^{\frac{1}{2}}} + \frac{1}{{\sqrt x }}} \right)dx} = \frac{2}{5}{x^{\frac{5}{2}}} + 2{x^{\frac{3}{2}}} + 2\sqrt x + C\).

c) \(\int {f\left( x \right)dx} = \int {\left( {\frac{1}{{\sqrt x }} + \frac{1}{x} - \frac{1}{{{x^2}}}} \right)dx} \)\( = 2\sqrt x + \ln \left| x \right| + \frac{1}{x} + C\).

Ta có \(F\left( x \right) = \int {f\left( x \right)dx} = \int {\frac{1}{{2x + 1}}dx} = \frac{1}{2}\ln \left| {2x + 1} \right| + C\).

Do \(F\left( 0 \right) = 2\) suy ra \(C = 2\) nên \(F\left( e \right) = \frac{1}{2}\ln \left( {2e + 1} \right) + 2 = \ln \sqrt {2e + 1} + 2\).

Hướng dẫn giải

a) \(f\left( x \right) = {8^x}{.2^{1 - 2x}} = {2^{3x}}{.2^{1 - 2x}} = {2^{1 + x}} = {2.2^x}\).

Suy ra \(\int {f\left( x \right)dx} = 2\int {{2^x}dx} = 2.\frac{{{2^x}}}{{\ln 2}} + C = \frac{{{2^{x + 1}}}}{{\ln 2}} + C\).

b) \(f\left( x \right) = {\left( {{3^x} - \frac{1}{{{3^x}}}} \right)^2} = {9^x} - 2 + {9^{ - x}}\).

Suy ra \(\int {f\left( x \right)dx} = \int {\left( {{9^x} - 2 + {9^{ - x}}} \right)dx} = \frac{{{9^x}}}{{\ln 9}} - 2x - \frac{1}{{{9^x}\ln 9}} + C = \frac{{{9^x}}}{{2\ln 3}} - 2x - \frac{1}{{{{2.9}^x}\ln 3}} + C\).

c) \(\int {f\left( x \right)dx} = \int {\left( {{e^{x + 1}} - \frac{e}{{{x^2}}}} \right)dx} = {e^{x + 1}} + \frac{e}{x} + C\).

Hướng dẫn giải

a) Ta có \(h\left( x \right) = \int {h'\left( x \right)dx} = \int {\frac{1}{x}dx} = \ln x + C\).

Vì \(h\left( 1 \right) = 2\) nên \(\ln 1 + C = 2 \Rightarrow C = 2\).

Do đó chiều cao của cây sau \(x\) năm \(\left( {1 \le x \le 11} \right)\) là \(h\left( x \right) = \ln x + 2\).

b) Ta có \(h\left( x \right) = 3 \Leftrightarrow \ln x + 2 = 3 \Leftrightarrow x = e \approx 2,72\) năm.

Vậy sau khoảng 2,72 năm thì cây cao 3 m.

a) \(\int\limits_1^2 {\frac{{{x^2} - 2x + 1}}{x}dx} = \int\limits_1^2 {\left( {x - 2 + \frac{1}{x}} \right)dx = \left. {\left( {\frac{{{x^2}}}{2} - 2x + \ln x} \right)} \right|} _1^2 = - \frac{1}{2} + \ln 2\).

b) \(\int\limits_1^4 {\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)dx} \)\( = \int\limits_1^4 {\left( {x - 4} \right)dx = \left. {\left( {\frac{{{x^2}}}{2} - 4x} \right)} \right|} _1^4 = - \frac{9}{2}\).

c) \[\int\limits_0^{2024} {{3^x}dx} = \left. {\frac{{{3^x}}}{{\ln 3}}} \right|_0^{2024} = \frac{{{3^{2024}}}}{{\ln 3}} - \frac{1}{{\ln 3}}\].

a) \(\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}x}}{{1 + \cos x}}dx} \)\( = \int\limits_0^{\frac{\pi }{2}} {\frac{{1 - {{\cos }^2}x}}{{1 + \cos x}}dx} \)\( = \int\limits_0^{\frac{\pi }{2}} {\left( {1 - \cos x} \right)dx} = \left. {\left( {x - \sin x} \right)} \right|_0^{\frac{\pi }{2}} = \frac{\pi }{2} - 1\).

b)\(\int\limits_0^{\frac{\pi }{4}} {\frac{{\cos 2x}}{{\cos x\left( {1 + \tan x} \right)}}dx} \)\( = \int\limits_0^{\frac{\pi }{4}} {\frac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{\cos x + \sin x}}dx} \)\( = \int\limits_0^{\frac{\pi }{4}} {\left( {\cos x - \sin x} \right)dx} = \left. {\left( {\sin x + \cos x} \right)} \right|_0^{\frac{\pi }{4}} = \sqrt 2 - 1\).

c) \(\int\limits_0^{\frac{\pi }{2}} {\left( {2{{\sin }^2}x + 3} \right)dx} \)\( = \int\limits_0^{\frac{\pi }{2}} {\left( {1 - \cos 2x + 3} \right)dx} \)\( = \int\limits_0^{\frac{\pi }{2}} {\left( {4 - \cos 2x} \right)dx} \)\( = \left. {\left( {4x - \frac{1}{2}\sin 2x} \right)} \right|_0^{\frac{\pi }{2}} = 2\pi \).