Question

[Quỳnh Như Trần Thị]

Bài tập: Tính.

b, \(\int\limits^{\dfrac{\pi}{6}}_0cos2xdx\)                                              d, \(\int\limits^2_1\dfrac{dx}{\left(2x-1\right)^2}\)

c, \(\int\limits^1_{-1}\left(2x+1\right)^3dx\)

Answer 1

[Nguyễn Việt Lâm]

\(b=\dfrac{1}{2}\int\limits^{\dfrac{\pi}{6}}_0cos2xd\left(2x\right)=\dfrac{1}{2}sin2x|^{\dfrac{\pi}{6}}_0=\dfrac{\sqrt{3}}{4}\)

\(c=\dfrac{1}{2}\int\limits^1_{-1}\left(2x+1\right)^3d\left(2x+1\right)=\dfrac{1}{2}.\dfrac{1}{4}\left(2x+1\right)^4|^1_{-1}=10\)

\(d=\dfrac{1}{2}\int\limits^2_1\dfrac{d\left(2x-1\right)}{\left(2x-1\right)^2}=-\dfrac{1}{2}.\dfrac{1}{\left(2x-1\right)}|^2_1=\dfrac{1}{3}\)

Answer 2

[Quỳnh Như Trần Thị]

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