Đề kiểm tra 1 tiết Chương 3 giải tích 12 Trường THPT Triệu Quang Phục năm 2017 - 2018 (Phần...

A. \(\int {\frac{{dx}}{{5x - 2}} = 5\ln \left| {5x - 2} \right| + C} \)

B. \(\int {\frac{{dx}}{{5x - 2}} =  - \frac{1}{2}\ln (5x - 2) + C} \)

C. \(\int {\frac{{dx}}{{5x - 2}} = \ln \left| {5x - 2} \right| + C} \)

D. \(\int {\frac{{dx}}{{5x - 2}} = \frac{1}{5}\ln \left| {5x - 2} \right| + C} \)

A. \(\int {f\left( x \right){\rm{d}}x}  =  - \cot x + C\)

B. \(\int {f\left( x \right){\rm{d}}x}  =   \tan x + C\)

C. \(\int {f\left( x \right){\rm{d}}x}  =  - \tan x + C\)

D. \(\int {f\left( x \right){\rm{d}}x}  =   \cot x + C\)

A. \(\int {2\sin xdx}  = {\sin ^2}x + C\)

B. \(\int {2\sin xdx}  = 2\cos x + C\)

C. \(\int {2\sin xdx}  =  - 2\cos x + C\)

D. \(\int {2\sin xdx}  = \sin 2x + C\)

A. \(I = \int\limits_0^3 {\sqrt u du} \)

B. \(I = \frac{1}{2}\int\limits_1^2 {\sqrt u du} \)

C. \(I = 2\int\limits_0^3 {\sqrt u du} \)

D. \(I = \int\limits_1^2 {\sqrt u du} \)

A. \(\int\limits_a^b {f(x)dx}  = F(b) + F(a)\)

B. \(\int\limits_a^b {F(x)dx}  = f(b) + f(a)\)

C. \(\int\limits_a^b {f(x)dx}  = F(b) - F(a)\)

D. \(\int\limits_a^b {F(x)dx}  = f(b) - f(a)\)

A. \(I=27\)

B. \(I=3\)

C. \(I=9\)

D. \(I=1\)

A. \(24\)

B. \(-7\)

C. \(-4\)

D. \(8\)

A. \(I = \frac{2}{{\ln 3}}\)

B. \(I = \frac{3}{{\ln 3}}\)

C. \(I=2\)

D. \(I = \frac{1}{4}\)

A. \(\int\limits_a^b {2({x^2} + 1)dx} \)

B. \(\int\limits_a^b {2\sqrt {{x^2} + 1} dx} .\)

C. \(\int\limits_a^b {2\pi ({x^2} + 1)dx} \)

D. \(\pi \int\limits_a^b {4({x^2} + 1)dx} \)

A. \(F(x) = {e^x} + {x^2} + \frac{3}{2}\)

B. \(F(x) = {e^x} + {x^2} + \frac{5}{2}\)

C. \(F(x) = {e^x} + {x^2} + \frac{1}{2}\)

D. \(F(x) = 2{e^x} + {x^2} - \frac{1}{2}\)

A. \(\int {f(ax + b)}  = \frac{1}{a}F(ax + b) + C\)

B. \(\int {f(ax + b)}  = aF(ax + b) + C\)

C. \(\int {f(ax + b)}  = F(ax + b) + C\)

D. \(\int {f(ax + b)}  = aF(x) + b + C\)

A. \(\frac{{{\pi ^2}}}{2}\)

B. \(\frac{{3{\pi ^2}}}{4} - 2\pi \)

C. \( - \frac{{3{\pi ^2}}}{4} + \pi \)

D. \(\frac{{{\pi ^2}}}{4}\)

A. \(a-b\)

B. \(-a-b\)

C. \(a+b\)

D. \(b-a\)

A. \(a=2\)

B. \(a=-2\)

C. \(a=1\)

D. \(a=0\)

A. \(\int {\cos 3xdx = \frac{{\sin 3x}}{3} + C} \)

B. \(\int {\cos 3xdx = 3\sin 3x + C} \)

C. \(\int {\cos 3xdx =  - \frac{{\sin 3x}}{3} + C} \)

D. \(\int {\cos 3xdx = \sin 3x + C} \)

A. \(\left| {\int\limits_{ - 2}^1 {({x^3} - x)dx} } \right|.\)

B. \(\int\limits_{ - 2}^1 {({x^3} - x)dx} .\)

C. \(\int\limits_{ - 1}^1 {\left| {{x^3} - x} \right|dx} .\)

D. \(\int\limits_{ - 2}^1 {\left| {{x^3} - x} \right|dx} .\)

A. \(\sin 2x + C\)

B. \( - \cos x - \sin x + C\)

C. \(\cos x + \sin x + C\)

D. \(\sin x - \cos x + C\)

A. \(3\)

B. \(1\)

C. \(-1\)

D. \(2\)

A. \(I = \frac{1}{4}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right|\)

B. \(I = \frac{1}{2}\ln \left| {\frac{{x + 2}}{{x - 2}}} \right|\)

C. \(I = \frac{1}{4}\ln \left| {\frac{{x + 2}}{{x - 2}}} \right|\)

D. \(I = \frac{1}{2}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right|\)

A. \(9\)

B. \(\frac{1}{3}\)

C. \(17\)

D. \(7\)