Bài tập 5 trang 127 SGK Giải tích 12

Bài tập 5 trang 127 SGK Giải tích 12

Câu a:

\(\int_{0}^{3}\frac{x}{\sqrt{1+x}}dx=\int_{0}^{3}\frac{x+1-1}{\sqrt{1+x}}dx\)

\(=\int_{0}^{3}\frac{x+1}{\sqrt{1+x}}dx-\int_{0}^{3}\frac{dx}{\sqrt{1+x}}\)

\(=\int_{0}^{3}(x+1)^{\frac{1}{2}}dx-\int_{0}^{3}(1+x)^{\frac{1}{2}}dx\)

\(= \int_{0}^{3}(x+1)^{\frac{1}{2}}d(x+1)-\int_{0}^{3}(1+x)^{\frac{1}{2}}d(1+x)\)

\(=\frac{2}{3}(x+1)^\frac{3}{2} \Bigg |^3_0-2(1+x)^{\frac{1}{2}}\Bigg |^3_0= \frac{2}{3}-2.4^\frac{1}{2}+2\)

\(=\frac{16}{3}-\frac{2}{3}-4+2=\frac{14}{3}-2=\frac{8}{3}\)

Câu b:

Đặt \(t=\sqrt[6]{x}\Rightarrow x=t^6\Rightarrow dx=6t^5 dt.\)

Khi x = 1 ⇒ t = 1

Khi x = 64 ⇒ t = 2

Do đó ta có: \(I=\int_{1}^{64}\frac{1+\sqrt{x}}{\sqrt[3]{x}}dx= \int_{1}^{2}\frac{1+t^3}{t^2}.6t^5dt =\int_{1}^{2}(6t^3+6t^6)dt\)

\(=\left [ \frac{3t^4}{2} +\frac{6t^7}{7} \right ] \Bigg |^2_1= \frac{3(16-1)}{2}+\frac{6(128-1)}{7}=\frac{1839}{14}.\)

Câu c:

Sử dụng phương pháp tính tích phân từng phần bằng cách đặt: 

\(\left\{\begin{matrix} u=x^2\\ dv=e^{3x}dx \end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2x dx\\ v=\frac{1}{3}e^{3x} \end{matrix}\right.\)

Ta có:

\(\int_{0}^{2}x^2 e^{3x}dx=\frac{1}{3}x^3.e^{3x} \bigg|_0^2-\frac{2}{3} \int_{0}^{2}x.e^{3x}dx=\frac{4}{3}e^6-\frac{2}{3}\int_{0}^{2}x.e^{3x}dx\)

Lại tiếp tục đặt: \(\left\{\begin{matrix} u_1=x^2\\ dv_1=e^{3x}dx \end{matrix}\right.\Rightarrow \left\{\begin{matrix} du_1=dx\\ v_1=\frac{1}{3}e^{3x} \end{matrix}\right.\)

Ta được:

\(\int_{0}^{2}x.e^{3x}dx=\frac{1}{3}x.e^{3x} \bigg|^2_0-\frac{1}{3} \int_{0}^{2}e^{3x}dx\)

\(=\frac{2}{3}e^6-\frac{1}{9}e^{3x}\bigg|_{0}^{2}=\frac{2}{3}e^6-\frac{1}{9}e^6+ \frac{1}{9}\)

Do đó: \(\int_{0}^{2}x^2.e^{3x}dx=\frac{4}{3}e^6-\frac{2}{3} \left ( \frac{2}{3}e^6-\frac{1}{9}e^6+\frac{1}{9} \right )\)

\(=\frac{4}{3}e^6-\frac{4}{9}e^6+\frac{2}{27}e^6-\frac{2}{27}= \frac{26}{27}e^6-\frac{2}{27}=\frac{2}{27}(13e^6-1)\)

Câu d:

Ta có:

\(\sqrt{1+sin2x}=\sqrt{(sinx+cosx)^2}= \sqrt{2cos^2(x-\frac{\pi }{4})}\)

\(= \sqrt{2}\left | cos\left ( x-\frac{\pi }{4} \right ) \right |\)

Mặt khác ta có:

\(cos\left ( x-\frac{\pi }{4} \right )\geq 0, \ \ \ \forall x\in \left [ 0;\frac{3\pi }{4} \right ]\)

và \(cos\left ( x-\frac{\pi }{4} \right )\leq 0, \ \ \ \forall x\in \left [\frac{3\pi }{4}; \pi \right ]\)

Do đó:

\(\int_{0}^{\pi }\sqrt{1+sin2x}dx=\sqrt{2}\int_{0}^{\pi }cos \left ( x-\frac{\pi }{4} \right ) \bigg| dx\)

\(=\sqrt{2}\int_{0}^{\frac{3\pi }{4}}cos\left ( x-\frac{\pi }{4} \right )dx- \sqrt{2}\int^{0}_{\frac{3\pi }{4}}cos\left ( x-\frac{\pi }{4} \right )dx\)

\(=\sqrt{2}sin\left ( x-\frac{\pi }{4} \right ) \Bigg|_0^{\frac{3\pi }{4}}- \sqrt{2}sin\left ( x-\frac{\pi }{4} \right )\Bigg|^0_{\frac{3\pi }{4}}\)

\(=\sqrt{2}\left ( 1+\frac{\sqrt{2}}{2} \right )-\sqrt{2} \left ( \frac{\sqrt{2}}{2} -1\right )= \sqrt{2}+\sqrt{2}=2\sqrt{2}\)

-- Mod Toán 12