Chương 3: NGUYÊN HÀM. TÍCH PHÂN VÀ ỨNG DỤNG - Hỏi đáp

Bài 4:

Để chứng minh $f(x,y)$ liên tục tại $(0;0)$ ta chứng minh \(\lim\limits _{x\to 0, y\to 0}f(x)=\lim\limits _{x\to 0, y\to 0}\frac{xy}{\sqrt{x^2+y^2}}=0\)

Thật vậy:

Với mọi \(\varepsilon >0\) ta chọn \(\delta=2\varepsilon\). Khi $\sqrt{x^2+y^2}< \delta$ thì ta luôn có:

\(|f(x)|=\frac{|xy|}{\sqrt{x^2+y^2}}\leq \frac{x^2+y^2}{2\sqrt{x^2+y^2}}=\frac{\sqrt{x^2+y^2}}{2}< \frac{\delta}{2}=\varepsilon\)

Theo định nghĩa giới hạn suy ra

\(\lim\limits_{x\to 0; y\to 0}f(x,y)=0=f(0,0)\) nên hàm liên tục tại $(0;0)$

Bài 4:

Thấy rằng:

\(\left\{\begin{matrix} z'_x=1-2x=0\\ z'_y=2-2y=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{1}{2}\\ y=1\end{matrix}\right.\)

Do đó hàm số có điểm dừng $M(\frac{1}{2}, 1)$

Tại $M(\frac{1}{2}, 1)$ ta thấy:

\(\left\{\begin{matrix} z''_{xx}=-2=A\\ z''_{xy}=0=B\\ z''_{yy}=-2=C\end{matrix}\right.\)

Ta thấy: $B^2-AC< 0$ và $A<0$ nên hàm $z$ có cực đại tại $(x,y)=(\frac{1}{2}, 0)$

Vậy $z_{cđ}=\frac{45}{4}$ tại $(x,y)=(\frac{1}{2},0)$

Đặt \(x=\frac{\sqrt{2}}{2}sint\Rightarrow dx=\frac{\sqrt{2}}{2}cost.dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=\frac{1}{2}\Rightarrow t=\frac{\pi}{4}\end{matrix}\right.\)

\(\int\limits^{\frac{1}{2}}_0f\left(\sqrt{1-2x^2}\right)dx=\frac{\sqrt{2}}{2}\int\limits^{\frac{\pi}{4}}_0f\left(cost\right).costdt=\frac{\sqrt{2}}{2}\int\limits^{\frac{\pi}{4}}_0f\left(cosx\right)cosxdx=\frac{7}{6}\)

\(\Rightarrow J=\int\limits^{\frac{\pi}{4}}_0f\left(cosx\right).cosx.dx=\frac{7\sqrt{2}}{6}\)

Đặt \(\left\{{}\begin{matrix}u=f\left(cosx\right)\\dv=cosx.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-sinx.f'\left(cosx\right)dx\\v=sinx\end{matrix}\right.\)

\(\Rightarrow J=sinx.f\left(cosx\right)|^{\frac{\pi}{4}}_0+\int\limits^{\frac{\pi}{4}}_0f'\left(cosx\right)sin^2x.dx=\frac{\sqrt{2}}{2}+I\)

\(\Rightarrow I=\frac{7\sqrt{2}}{6}-\frac{\sqrt{2}}{2}=\frac{2\sqrt{2}}{3}\)

Lời giải:

Bổ sung điều kiện $f(x)$ liên tục trên đoạn $[0;1]$

Ta có:

ĐKĐB $\Rightarrow \int ^1_04f(x)dx+\int ^1_03f(1-x)dx=\int ^1_0\sqrt{1-x}dx$

$\Leftrightarrow 4\int ^1_0f(x)dx+3\int ^1_0f(1-x)dx=\frac{2}{3}$

Mà:

$\int ^1_0f(1-x)dx=-\int ^1_0f(1-x)d(1-x)=-\int ^0_1f(x)dx=\int ^1_0f(x)dx$

Do đó:

$7\int ^1_0f(x)dx=\frac{2}{3}$

$\Rightarrow \int ^1_0f(x)dx=\frac{2}{21}$

Chọn \(f\left(x\right)=x^3+ax^2+bx+c\)

\(2f\left(x^2\right)+f'\left(x\right)=2x^6+7x^2+2\)

\(\Leftrightarrow2x^6+2ax^4+2bx^2+c+3x^2+2ax+b=2x^6+7x^2+2\)

\(\Leftrightarrow2ax^4+\left(2b+3\right)x^2+2ax+b+c=7x^2+2\)

Đồng nhất 2 vế ta được: \(\left\{{}\begin{matrix}a=0\\2b+3=7\\b+c=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c=0\\b=2\end{matrix}\right.\)

\(\Rightarrow f\left(x\right)=x^3+2x\Rightarrow f\left(1\right)=3\)

k/ \(I=\int x.sin\left(2x+1\right)dx\)

Đặt \(\left\{{}\begin{matrix}u=x\\dv=sin\left(2x+1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-\frac{1}{2}cos\left(2x+1\right)\end{matrix}\right.\)

\(\Rightarrow I=-\frac{1}{2}x.cos\left(2x+1\right)+\frac{1}{2}\int cos\left(2x+1\right)dx\)

\(=-\frac{1}{2}x.cos\left(2x+1\right)+\frac{1}{4}sin\left(2x+1\right)+C\)

l/ \(I=\int\left(1-2x\right)e^{3x}dx\)

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

\(\Rightarrow I=\frac{1}{3}\left(1-2x\right)e^{3x}+\frac{2}{3}\int e^{3x}dx=\frac{1}{3}\left(1-2x\right)e^{3x}+\frac{2}{9}e^{3x}+C\)

h/ \(I=\int e^xcosx.dx\)

Đặt \(\left\{{}\begin{matrix}u=cosx\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-sinx.dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I=e^x.cosx+\int e^xsinx.dx\)

Xét \(J=\int e^xsinx.dx\)

Đặt \(\left\{{}\begin{matrix}u=sinx\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=cosxdx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow J=e^x.sinx-\int e^x.cosxdx=e^xsinx-I\)

\(\Rightarrow I=e^xcosx+e^xsinx-I\)

\(\Rightarrow2I=e^x\left(sinx+cosx\right)\Rightarrow I=\frac{1}{2}e^x\left(sinx+cosx\right)+C\)

\(\int\limits^{\frac{\pi}{2}}_0\frac{cosx}{1+sin^2x}dx=\int\limits^{\frac{\pi}{2}}_0\frac{d\left(sinx\right)}{1+sin^2x}=arctan\left(sinx\right)|^{\frac{\pi}{2}}_0=\frac{\pi}{4}\)

\(I=\int\limits^2_1\frac{x\left(3x-\sqrt{9x^2-1}\right)}{\left(3x+\sqrt{9x^2-1}\right)\left(3x-\sqrt{9x^2-1}\right)}dx=\int\limits^2_1x\left(3x-\sqrt{9x^2-1}\right)dx\)

\(=\int\limits^2_13x^2dx-\frac{1}{18}\int\limits^2_1\sqrt{9x^2-1}.d\left(9x^2-1\right)\)

\(=x^3|^2_1-\frac{1}{27}\sqrt{\left(9x^2-1\right)^3}|^2_1=7-\frac{1}{27}\left(35\sqrt{35}-16\sqrt{2}\right)\)

\(f\left(x\right)=\int sin^4xdx=\int\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2dx\)

\(=\frac{1}{4}\int\left(1-2cos2x+cos^22x\right)dx=\frac{1}{4}\int\left(\frac{3}{2}-2cos2x+\frac{1}{2}cos4x\right)dx\)

\(=\frac{1}{4}\left(\frac{3}{2}x-sin2x+\frac{1}{8}sin4x\right)+C\)

\(f\left(0\right)=0\Rightarrow\frac{1}{4}\left(0-0+0\right)+C=0\Rightarrow C=0\)

\(\Rightarrow\int\limits^{\frac{\pi}{2}}_0f\left(x\right)dx=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_0\left(\frac{3}{2}x-sin2x+\frac{1}{8}sin4x\right)dx\)

\(=\frac{1}{4}\left(\frac{3}{4}x^2+\frac{1}{2}cos2x-\frac{1}{32}cos4x\right)|^{\frac{\pi}{2}}_0\)

\(=\frac{3\pi^2-16}{64}\)

Tung độ giao điểm: \(y=4y-y^2\Rightarrow\left\{{}\begin{matrix}y=0\Rightarrow x=0\\y=3\Rightarrow x=3\end{matrix}\right.\)

\(x=4y-y^2=4-\left(y-2\right)^2\Rightarrow\left(y-2\right)^2=4-x\)

\(\Rightarrow y=2+\sqrt{4-x}\) với \(y\ge2\)

\(y=2-\sqrt{4-x}\) với \(y\le2\)

Thể tích:

\(V=\pi\int\limits^3_0x^2dx+\pi\int\limits^4_3\left(2+\sqrt{4-x}\right)^2dx-\pi\int\limits^4_0\left(2-\sqrt{4-x}\right)^2dx=\frac{27\pi}{2}\)

Sxq = \(\pi rl\)

Chắc chắn là bạn ghi thiếu đề hoặc sai đề :)

Tích phân đúng như thế này thì ko tính được

e) \(I_4=\int_0^{\pi}\sqrt{\frac{1+\cos^2x}{2}}dx=\int_0^{\pi}\sqrt{\cos^2x}dx=\int_0^{\pi}\left|\cos x\right|dx\)

Mặt khác ta có \(\left|\cos x\right|=\cos x,0\le x\le\frac{\pi}{2}\) và \(\left|\cos x\right|=-\cos x,\frac{\pi}{2}\le x\le\pi\)

Do đó :

\(I_4=\int_0^{\frac{\pi}{2}}\cos xdx-\int_{^{\frac{\pi}{2}}}^{\pi}\cos xdx\)

    \(=\sin x^{\frac{\pi}{2}}_0-\sin x^{\pi}_{\frac{\pi}{2}}\)

    \(=\left(1-0\right)-\left(0-1\right)=2\)

d) Áp dụng công thức tích phân từng phần \(\int_a^budv=uv^b_a-\int_a^bvdu\) cho \(u=x+3;dv=\sin axdx\)

Ta tìm được \(I_5=\left(-\frac{x+3}{a}\cos ax+\frac{1}{a^2}\sin ax\right)_0^{\frac{\pi}{2a}}\)

                        \(=\frac{1+3a}{a^2}\)

Pt hoành độ giao điểm:

\(1+sinx=1\Rightarrow\left[{}\begin{matrix}x=0\\x=\pi\end{matrix}\right.\) ; \(\left[{}\begin{matrix}x=0\Rightarrow y=1\\x=\frac{\pi}{2}\Rightarrow y=2\end{matrix}\right.\)

\(sinx=y-1\Rightarrow\left[{}\begin{matrix}x=\pi-arcsin\left(y-1\right)\\x=arcsin\left(y-1\right)\end{matrix}\right.\)

\(V=\pi\int\limits^2_1\left(\pi-arcsin\left(y-1\right)\right)^2dy-\pi\int\limits^2_1arcsin^2\left(y-1\right)dy\)

\(=\pi\int\limits^1_0\left(\pi^2-2\pi.arcsint+arcsin^2t\right)dt-\pi\int\limits^1_0arcsin^2t.dt\) (với \(t=y-1\))

\(=\pi\int\limits^1_0\pi^2dt-2\pi^2\int\limits^1_0arcsintdt\)

\(=\pi^3-2\pi^2\left(t.arcsint+\sqrt{1-t^2}\right)|^1_0\)

\(=\pi^3-2\pi^2\left(\frac{\pi}{2}-1\right)=2\pi^2\)

\(I=\int\limits^e_1\frac{1}{t}dt+\int\limits^e_1m.lnt.d\left(lnt\right)=lnt|^e_1+\frac{m}{2}.ln^2t|^e_1\)

\(=1+\frac{m}{2}=0\Rightarrow m=-2\)

\(\Rightarrow\) Đáp án D đúng