Trắc nghiệm Tích phân có đáp án (P1) (Thông hiểu)

A. I = 0

B. I = -1

C. I = 1

D. I = 2

A. 3

B. 30

C. -3

D. -30

A. a = 2

B. $a=\frac{1}{2}$

C. $a=\sqrt{2}$

D. a = 4

A. $I=-\frac{1}{4}{\pi }^4$

B. $I=-{\pi }^4$

C. $I=0$

D. $I=-\frac{1}{4}$

A. $I=2{\int }_8^9\sqrt{u}du$

B. $I=\frac{1}{2}{\int }_8^9\sqrt{u}du$

C. $I=2{\int }_9^8\sqrt{u}du$

D. $I={\int }_8^9\sqrt{u}du$

A. $I=\left(\frac{u^3}{3}+u\right){|}_1^2$

B. $I=\frac{4}{3}\left(u^3+u\right){|}_1^2$

C. $I=\frac{1}{3}\left(\frac{u^3}{3}+u\right){|}_1^2$

D. $I=2\left(\frac{u^3}{3}+u\right){|}_1^2$

A. f(x) = cos3x

B. f(x) = sin3x

C. $f\left(x\right)=\cos \left(\frac{x}{4}+\frac{\pi }{2}\right)$

D. $f\left(x\right)=\sin \left(\frac{x}{4}+\frac{\pi }{2}\right)$

A. $\frac{1}{2}\ln \frac{1}{3}$

B. $2\ln 3$

C. $\ln \sqrt{3}$

D. $2\ln \frac{1}{3}$

A. $\frac{2\ln 2}{3}$

B. $-\frac{2\ln 2}{3}$

C. -2ln2

D. 2ln2

A. $\left(\begin{array}{c}m=1\\ m=-2\end{array}\right)$

B. $\left(\begin{array}{c}m=1\\ m=2\end{array}\right)$

C. $\left(\begin{array}{c}m=-1\\ m=2\end{array}\right)$

D. $\left(\begin{array}{c}m=-1\\ m=-2\end{array}\right)$

A. $I=\pi$

B. $I=\frac{\pi }{6}$

C. $I=\frac{\pi }{3}$

D. $I=\frac{\pi }{2}$

A. a = 1

B. $a=-\frac{1}{3}$

C. a = 2

D. a = -2

A. $I=-{\int }_{\sqrt{2}}^{\frac{2}{\sqrt{3}}}\frac{t^2}{t^2-1}dt$

B. $I=-{\int }_2^3\frac{t^2}{t^2+1}dt$

C. $I={\int }_{\sqrt{2}}^{\frac{2}{\sqrt{3}}}\frac{t^2}{t^2-1}dt$

D. $I=-{\int }_{2\sqrt{2}}^3\frac{t}{t^2+1}dt$

A. $I={\int }_1^2\frac{4\left(t^2-1\right)}{3}dt$

B. $I={\int }_1^2\left(t^2-1\right)dt$

C. $I={\int }_1^2\frac{4\left(t^2-1\right)}{5}dt$

D. $I={\int }_1^2\frac{\left(t^2-1\right)}{3}dt$

A. $f\left(x\right)=-\frac{{\pi }^x}{\mathrm{ln\pi }}$

B. $f\left(x\right)=\frac{{\pi }^x}{\mathrm{ln\pi }}$

C. $f\left(x\right)={\pi }^x\mathrm{ln\pi }$

D. $f\left(x\right)=-{\pi }^x\mathrm{ln\pi }$