Bài tập 1 trang 112 SGK Giải tích 12

Bài tập 1 trang 112 SGK Giải tích 12

Câu a:

Đặt \(u=1-x\) ta có \(du=-dx\)

Khi \(x=-\frac{1}{2}\) thì \(u=\frac{3}{2}\); khi \(x=\frac{1}{2}\) thì \(u=\frac{1}{2}\). Do đó:

\(\int_{\frac{-1}{2}}^{\frac{1}{2}}\sqrt[3]{ (1-x)^{2}}dx\) \(=-\int_{\frac{3}{2}}^{\frac{1}{2}} \sqrt[3]{u^2} du = \int^{\frac{3}{2}}_{\frac{1}{2}} u^{\frac{3}{2}}du =\frac{3}{5}u^{\frac{5}{3}} \bigg|_{\frac{1}{2}}^{\frac{3}{2}}\)

 \(=\frac{3}{5} u\sqrt[3]{u^2} \bigg| ^{\frac{3}{2}}_{\frac{1}{2}}= \frac{3}{5} \left ( \frac{3}{2}\sqrt[3]{\frac{9}{4}}- \frac{1}{2}\sqrt[3]{\frac{1}{4}} \right )=\frac{3}{10\sqrt[3]{4}}(3\sqrt[3]{9}-1)\)

Câu b:

Đặt \(u=\frac{\pi }{4}-x\) ta có \(du=-dx\)

Khi x = 0 thì \(u=\frac{\pi }{4};\) khi \(x=\frac{\pi }{2}\) thì \(u=- \frac{\pi }{4}\). Do đó:

 \(\int_{0}^{\frac{\pi}{2}}sin(\frac{\pi}{4}-x)dx =-\int_{\frac{\pi}{4}}^{-\frac{\pi}{4}}sinu. du\)

\(=-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}sin u. du = -cos u \bigg|_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\)

\(=-\left ( cos \frac{\pi }{4} -cos \left ( -\frac{\pi }{4} \right ) \right )=0\)

Vậy \(\int_{0}^{\frac{\pi }{2}} sin \left ( \frac{\pi }{4} -x \right )dx = 0.\)

Câu c:

Ta có:

\(\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}\). Do đó:

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

\(=\int_{\frac{1}{2}}^{2}\frac{dx}{x}-\int_{\frac{1}{2}}^{2}\frac{d(x+1)}{x+1}= ln \left | x \right | \bigg|^2_{\frac{1}{2}}-ln \left | x+1 \right | \bigg|^2_{\frac{1}{2}}\)

\(=ln2 -ln\frac{1}{2}-ln3-ln\frac{3}{2}=ln2.\)

Câu d:

\(\int_{0}^{2}x(x+1)^2dx=\int_{0}^{2}(x^2+2x^2+x)dx\)

\(= \left ( \frac{x^4}{4}+\frac{2}{3}x^3+\frac{1}{2}x^2 \right ) \Bigg| ^2_0= 4+\frac{16}{3}+2=\frac{34}{3}\)

Câu e:

Đặt u = x + 1 ta có du = dx và x = u - 1

Khi \(x=\frac{1}{2}\) thì \(u=\frac{3}{2}\); khi \(x=2\) thì \(u=3\). Do đó:

\(\int_{\frac{1}{2}}^{2} \frac{1-3x}{(x+1)^2}dx=\int_{\frac{3}{2}}^{3} \frac{1-3(u-1)}{u^2}du=\int_{\frac{3}{2}}^{3}\frac{4-3u}{u^2}du\)

\(=4\int_{\frac{3}{2}}^{3}-3\int_{\frac{3}{2}}^{3}\frac{du}{u}= -\frac{4}{u} \Bigg |^3_{\frac{3}{2}}-3ln .u\Bigg |^3_{\frac{3}{2}}\)

\(=-\left ( \frac{4}{3} - \frac{4}{\frac{3}{2}}\right )-3 \left ( ln3-ln\frac{3}{2} \right )=\frac{4}{3}-3ln2\)

Câu g: 

Ta có: \(sin3x . cos5x =\frac{1}{2}(sin8x-sin2x)\)

Do đó:

\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin3x. cos5x dx =\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (sin8x - sin2x)dx\)

\(=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin8x dx -\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin 2x dx\)

\(=-\frac{1}{16}cos 8x \Bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\frac{1}{4} cos2x\Bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\)

\(=-\frac{1}{16} \left [ cos4\pi -cos(-4\pi) \right ]+ \frac{1}{4}\left [ cos \pi - cos(-\pi) \right ]\)

\(=-\frac{1}{16}(1-1)+\frac{1}{4}(-1+1)=0\)

-- Mod Toán 12

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