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