Bài tập 2 trang 112 SGK Giải tích 12

Câu a:

Ta có \(|1-x|=\left\{\begin{matrix} 1-x \ voi \ 0\leq x\leq 1\\ x-1 \ voi \ 1\leq x\leq 2 \end{matrix}\right.\)

Do đó:

\(\int_{0}^{2}\left | 1-x \right |dx=\int_{0}^{2}(1-x)dx+\int_{0}^{2}(x-1)dx\)

\(=\left ( x-\frac{x^2}{2} \right ) \Bigg|^1_0+ \left ( \frac{x^2}{2} -x \right ) \Bigg|^2_1=\frac{1}{2}+\frac{1}{2}=1\)

Câu b:

\(\int_{0}^{\frac{\pi}{2}}sin^{2}x dx\) =\(\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(1-cos2x)dx=\frac{1}{2}\left ( x-\frac{1}{2}sin2x \right)|_{0}^{\frac{\pi }{2}}=\frac{\pi }{4}\)

Câu c:

\(\int_{0}^{ln2} \frac{e^{2x+1}+1}{e^x}dx=\int_{0}^{ln2} (e^{x+1}+e^{-x})dx=\int_{0}^{ln2}e^{x+1}dx+\int_{0}^{ln2}e^{-x}dx\)

\(= e^{x+1}\Bigg |^{ln2}_0-e^{-x}\Bigg |^{ln2}_0\)

\(=e^{ln2+1}-e^1-e^{-ln2}+1=e^{ln2+1}-\frac{1}{e^{ln2}}-(e-1)= e+\frac{1}{2}.\)

Câu d:

\(\int_{0}^{\pi }sin2x.cos^2xdx=2\int_{0}^{\pi }sinx.cos^3xdx\)

Đặt \(u = \cos x \Rightarrow du =  - \sin xdx\)

Đổi cận: x=0 ta có u=1, \(x = \pi \) ta có u=-1

Do đó: \(2\int\limits_0^\pi  {\sin x.{{\cos }^3}xdx}  =  - 2\int\limits_1^{ - 1} {{u^3}du}  = 2\int\limits_{ - 1}^1 {{u^3}du}  = \left. {\frac{1}{2}{u^4}} \right|_{ - 1}^1 = 0\)

-- Mod Toán 12 HỌC247

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