Tính tích phân sau I = ∫ 0 1 1 − x 2 d x

Chọn B.

Có $\underset{0}{\overset{m}{\int }}(2x-4)\text{d}x={\left.\right|}_{\left(x^2-4x\right)}^o=m^2-4m$

Vậy $\underset{0}{\overset{m}{\int }}(2x-4)\text{d}x=5\iff m^2-4m-5=0\iff \left[\begin{array}{l}m=-1\\ m=5\end{array}\right.$

Chọn A.

Đặt $u=x^2,\text{d}v=\sin x\text{d}x\Rightarrow \text{d}u=2x\text{d}x,v=-\cos x$

Khi đó $I=\underset{0}{\overset{\pi }{\int }}{x}^2\sin x\text{d}x={\left[-x^2\cos x\right]}_0^{\pi }+2\underset{0}{\overset{\pi }{\int }}x\cos x\text{d}x={\pi }^2+2K$

Chọn B.

Đặt $x=2\sin t\Rightarrow \text{d}x=2\cos t\text{d}t$

Đổi cận: $x=1\Rightarrow t=\frac{\pi }{6};x=0\Rightarrow t=0$

Khi đó: $\underset{0}{\overset{1}{\int }}\frac{\text{d}x}{\sqrt{4-x^2}}=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{2\cos t\text{d}t}{\sqrt{4-4{\sin }^{2}t}}=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{2\cos t\text{d}t}{2\sqrt{1-{\sin }^{2}t}}=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\cos t\text{d}t}{\sqrt{{\cos }^{2}t}}=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\text{d}t$

Chọn D.

Đặt $u=\ln x,\text{d}v=\left(2x-1\right)\text{d}x$ , suy ra $\text{d}u=\frac{1}{x}\text{d}x,v=x^2-x$

Chọn B.

Đặt $x=-t\Rightarrow \underset{\frac{-\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=\underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}f(-t)(-dt)=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f(-t)dt=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f(-x)dx$