Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2a. Đề sai, nhìn biểu thức \(\dfrac{f'\left(x\right)}{f'\left(x\right)}dx\) là thấy
2b. Đồ thị hàm số không cắt Ox trên \(\left(0;1\right)\) nên diện tích cần tìm:
\(S=\int\limits^1_0\left(x^4-5x^2+4\right)dx=\dfrac{38}{15}\)
3a. Phương trình (P) theo đoạn chắn:
\(\dfrac{x}{4}+\dfrac{y}{-1}+\dfrac{z}{-2}=1\)
3b. Câu này đề sai, đề cho mặt phẳng (Q) rồi thì sao lại còn viết pt mặt phẳng (Q) nữa?
sorry thầy em xin sửa lại câu 3 b là
b) trong không gian Oxyz cho mặt phẳng (Q): 3x-y-2z+1=0.Viết phương trình mặt phẳng (P) song song với mặt phẳng (Q) và đi qua điểm M(0;0;1)
\(f\left(0\right)=-1\Rightarrow f'\left(0\right)+2=0\Leftrightarrow f'\left(0\right)=-2\)
\(\int\limits^1_0f\left(x\right)dx=\int\limits^1_0\dfrac{f'\left(x\right)-x.e^{3x}}{2}dx=\dfrac{1}{2}\int\limits^1_0f'\left(x\right)dx-\dfrac{1}{2}\int\limits^1_0x.e^{3x}dx=\dfrac{1}{2}f\left(x\right)|^1_0-\dfrac{1}{2}\int\limits^1_0xe^{3x}dx\)
\(I_1=\int xe^{3x}dx\)
\(\left\{{}\begin{matrix}u=x\\dv=e^{3x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{3}e^{3x}\end{matrix}\right.\)
\(\Rightarrow I_1=\dfrac{1}{3}xe^{3x}-\dfrac{1}{3}\int e^{3x}dx=\dfrac{1}{3}xe^{3x}-\dfrac{1}{9}e^{3x}\)
\(\Rightarrow I=\dfrac{1}{2}f\left(1\right)-\dfrac{1}{2}f\left(0\right)-\dfrac{1}{2}\left(\dfrac{1}{3}xe^{3x}-\dfrac{1}{9}e^{3x}\right)|^1_0\)
Èo, tắc chỗ f(1) rồi, vậy đành phải biến đổi để tìm f(x) luôn vậy, hmm
Thử nhân 2 vế với \(e^{2x}\) xem nào:
\(e^{2x}f'\left(x\right)-2e^{2x}f\left(x\right)=x.e^{5x}\Leftrightarrow\left(e^{2x}.f\left(x\right)\right)'=x.e^{5x}\)
Lay nguyen ham 2 ve:
\(e^{2x}.f\left(x\right)=\int x.e^{5x}dx\)
\(\left\{{}\begin{matrix}x=u\\dv=e^{5x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}dx=du\\v=\dfrac{1}{5}e^{5x}\end{matrix}\right.\)
\(\Rightarrow e^{2x}.f\left(x\right)=\int x.e^{5x}dx=\dfrac{1}{5}x.e^{5x}-\dfrac{1}{5}\int e^{5x}dx=\dfrac{1}{5}xe^{5x}-\dfrac{1}{25}e^{5x}+C\)
\(f\left(0\right)=-1\Leftrightarrow f\left(0\right)=-\dfrac{1}{25}+C=-1\Leftrightarrow C=-\dfrac{24}{25}\)
\(\Rightarrow f\left(x\right)=\dfrac{\dfrac{1}{5}xe^{5x}-\dfrac{1}{25}e^{5x}-\dfrac{24}{25}}{e^{2x}}\)
Vậy là xong rồi \(\Rightarrow f\left(1\right)=...\) , thay vô \(I=\dfrac{1}{2}f\left(1\right)-\dfrac{1}{2}.\left(-1\right)-\dfrac{1}{2}\left(\dfrac{1}{3}xe^{3x}-\dfrac{1}{9}e^{3x}\right)|^1_0\) là được nha :)
Nguyên tắc:
\(g\left(x\right).f'\left(x\right)+h\left(x\right).f\left(x\right)=p\left(x\right)\)
Đầu tiên luôn biến đổi để \(f'\left(x\right)\) đứng riêng biệt 1 mình:
\(\Rightarrow f'\left(x\right)+\dfrac{h\left(x\right)}{g\left(x\right)}.f\left(x\right)=\dfrac{p\left(x\right)}{g\left(x\right)}\) (1)
Cần thêm/bớt, nhân/chia sao cho biến về dạng:
\(\left[u\left(x\right).f\left(x\right)\right]'=q\left(x\right)\)
\(\Leftrightarrow f'\left(x\right).u\left(x\right)+u'\left(x\right).f\left(x\right)=q\left(x\right)\)
\(\Leftrightarrow f'\left(x\right)+\dfrac{u'\left(x\right)}{u\left(x\right)}.f\left(x\right)=\dfrac{q\left(x\right)}{u\left(x\right)}\)
Chỉ quan tâm vế trái, khi đó ta sẽ thấy hàm đằng trước \(f\left(x\right)\) chính là \(\dfrac{u'\left(x\right)}{u\left(x\right)}\)
Đồng nhất \(\Rightarrow\dfrac{u'\left(x\right)}{u\left(x\right)}=-2\)
Lấy nguyên hàm 2 vế \(\Rightarrow ln\left|u\left(x\right)\right|=-2x\Rightarrow u\left(x\right)=e^{-2x}\)
Do đó, ở bài toán ban đầu ta cần nhân 2 vế của (1) với \(u\left(x\right)=e^{-2x}\) nghĩa là:
\(f'\left(x\right)-2f\left(x\right)=x.e^{3x}\Leftrightarrow e^{-2x}.f'\left(x\right)-2e^{-2x}.f\left(x\right)=x.e^x\)
\(\Leftrightarrow\left[e^{-2x}.f\left(x\right)\right]'=x.e^x\)
Nguyên hàm 2 vế: \(\Rightarrow e^{-2x}.f\left(x\right)=\left(x-1\right)e^x+C\)
Thay \(x=0\Rightarrow1.f\left(0\right)=-1+C\Rightarrow C=0\)
\(\Rightarrow e^{-2x}.f\left(x\right)=\left(x-1\right)e^x\Rightarrow f\left(x\right)=\left(x-1\right)e^{3x}\)
\(\Rightarrow I=\int\limits^1_0\left(x-1\right)e^{3x}dx=...\)
Câu 1:
\(\int\limits^3_0\left(f'\left(x\right)+1\right)\sqrt{x+1}dx=\int\limits^3_0f'\left(x\right)\sqrt{x+1}dx+\int\limits^3_0\sqrt{x+1}dx\)
\(=\int\limits^3_0f'\left(x\right)\sqrt{x+1}dx+\frac{14}{3}=\frac{302}{15}\Rightarrow\int\limits^1_0f'\left(x\right)\sqrt{x+1}dx=\frac{232}{15}\)
Ta có:
\(I=\int\limits^3_0\frac{f\left(x\right)dx}{\sqrt{x+1}}\)
Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=\frac{dx}{\sqrt{x+1}}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=2\sqrt{x+1}\end{matrix}\right.\)
\(\Rightarrow I=2f\left(x\right)\sqrt{x+1}|^3_0-2\int\limits^3_0f'\left(x\right)\sqrt{x+1}dx\)
\(=4f\left(3\right)-2f\left(0\right)-2.\frac{232}{15}\)
\(=2\left(2f\left(3\right)-f\left(0\right)\right)-\frac{464}{15}=36-\frac{464}{15}=\frac{76}{15}\)
Câu 2:
\(I_1=\int\limits^3_1\frac{xf'\left(x\right)}{x+1}dx=0\)
Đặt \(\left\{{}\begin{matrix}u=\frac{x}{x+1}\\dv=f'\left(x\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{1}{\left(x+1\right)^2}dx\\v=f\left(x\right)\end{matrix}\right.\)
\(\Rightarrow I_1=\frac{xf\left(x\right)}{x+1}|^3_1-\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}=\frac{3.3}{3+1}-\frac{1.3}{1+1}-\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx=\frac{3}{4}-\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx=0\)
\(\Rightarrow\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx=\frac{3}{4}\)
Ta có:
\(I=\int\limits^3_1\frac{f\left(x\right)+lnx}{\left(x+1\right)^2}dx=\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx+\int\limits^3_1\frac{lnx}{\left(x+1\right)^2}dx=\frac{3}{4}+I_2\)
Xét \(I_2=\int\limits^3_1\frac{lnx}{\left(x+1\right)^2}dx\Rightarrow\) đặt \(\left\{{}\begin{matrix}u=lnx\\dv=\frac{1}{\left(x+1\right)^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{dx}{x}\\v=\frac{-1}{x+1}\end{matrix}\right.\)
\(\Rightarrow I_2=\frac{-lnx}{x+1}|^3_1+\int\limits^3_1\frac{dx}{x\left(x+1\right)}=-\frac{1}{4}ln3+\int\limits^1_0\left(\frac{1}{x}-\frac{1}{x+1}\right)dx\)
\(=-\frac{1}{4}ln3+ln\left(\frac{x}{x+1}\right)|^3_1=-\frac{1}{4}ln3+ln\frac{3}{4}-ln\frac{1}{2}=\frac{3}{4}ln3-ln2\)
\(\Rightarrow I=\frac{3}{4}+\frac{3}{4}ln3-ln2\)
1/ \(4\int\limits^5_{-3}f'\left(x\right)dx=4f\left(x\right)|^5_{-3}=4\left[f\left(5\right)-f\left(-3\right)\right]=4.\left(9-1\right)=32\)
2/ \(\int\left(2x+1\right)e^xdx\)
\(\left\{{}\begin{matrix}u=2x+1\\dv=e^xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow\int\left(2x+1\right)e^xdx=\left(2x+1\right)e^x-2\int e^xdx=\left(2x+1\right)e^x-2e^x\)
P/s: Bạn tự thay cận vô nhé!
Xét \(I=\int\limits^1_0x^2f\left(x\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=x^2dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{3}x^3.f\left(x\right)|^1_0-\dfrac{1}{3}\int\limits^1_0x^3.f'\left(x\right)dx=-\dfrac{1}{3}\int\limits^1_0x^3f'\left(x\right)dx\)
\(\Rightarrow\int\limits^1_0x^3f'\left(x\right)dx=-1\)
Lại có: \(\int\limits^1_0x^6.dx=\dfrac{1}{7}\)
\(\Rightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+14\int\limits^1_0x^3.f'\left(x\right)dx+49.\int\limits^1_0x^6dx=0\)
\(\Rightarrow\int\limits^1_0\left[f'\left(x\right)+7x^3\right]^2dx=0\)
\(\Rightarrow f'\left(x\right)+7x^3=0\)
\(\Rightarrow f'\left(x\right)=-7x^3\)
\(\Rightarrow f\left(x\right)=\int-7x^3dx=-\dfrac{7}{4}x^4+C\)
\(f\left(1\right)=0\Rightarrow C=\dfrac{7}{4}\)
\(\Rightarrow I=\int\limits^1_0\left(-\dfrac{7}{4}x^4+\dfrac{7}{4}\right)dx=...\)
Ta có: 2 ∫ 1 [2 + f(x)] dx = 2 ∫ 1 [2 + x^2] dx = 2 [2x + (1/3)x^3]1→1 = 2 [(2+1/3) - (2+1/3)] = 0 Vậy đáp án là D. 7/3.
Xét \(I=\int\limits^1_0x.f\left(3x\right)dx\)
Đặt \(3x=u\Rightarrow dx=\dfrac{1}{3}du\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=0\\x=1\Rightarrow u=3\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{9}\int\limits^3_0u.f\left(u\right)du=\dfrac{1}{9}\int\limits^3_0x.f\left(x\right)dx=1\)
\(\Rightarrow J=\int\limits^3_0x.f\left(x\right)dx=9\)
Xét J, đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=x.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\dfrac{x^2}{2}\end{matrix}\right.\)
\(\Rightarrow J=\dfrac{x^2}{2}.f\left(x\right)|^3_0-\dfrac{1}{2}\int\limits^3_0x^2.f'\left(x\right)dx=\dfrac{9}{2}-\dfrac{1}{2}\int\limits^3_0x^2.f'\left(x\right)dx\)
\(\Rightarrow\int\limits^3_0x^2.f'\left(x\right)dx=9-2J=-9\)
Câu 1:
\(\left(x+2\right)f\left(x\right)+x\left(x+1\right)f'\left(x\right)=x\left(x+1\right)\)
\(\Leftrightarrow x\left(x+2\right)f\left(x\right)+x^2\left(x+1\right)f'\left(x\right)=x^2\left(x+1\right)\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{\left(x+1\right)^2}f\left(x\right)+\frac{x^2}{x+1}f'\left(x\right)=\frac{x^2}{x+1}\)
\(\Leftrightarrow\left(\frac{x^2}{x+1}f\left(x\right)\right)'=\frac{x^2}{x+1}=x-1+\frac{1}{x+1}\)
Lấy nguyên hàm 2 vế:
\(\Leftrightarrow\frac{x^2}{x+1}.f\left(x\right)=\frac{x^2}{2}-x+ln\left|x+1\right|+C\)
Thay \(x=1\Rightarrow ln2+\frac{1}{2}=\frac{1}{2}-1+ln2+C\Rightarrow C=1\)
\(\Rightarrow\frac{x^2}{x+1}f\left(x\right)=\frac{x^2}{2}-x+ln\left|x+1\right|+1\)
Thay \(x=2\Rightarrow\frac{4}{3}f\left(2\right)=ln3+1\Rightarrow f\left(2\right)=\frac{3}{4}ln+\frac{3}{4}\Rightarrow T=-\frac{3}{16}\)
Câu 2:
\(I_1=\int\limits^2_0f\left(x\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I_1=x.f\left(x\right)|^2_0-\int\limits^2_0x.f'\left(x\right)dx=2-\int\limits^2_0x.f'\left(x\right)dx\)
Mà \(I_1=2\)\(\Rightarrow I_2=\int\limits^2_0x.f'\left(x\right)dx=-2\)
Đặt \(2x=t\Rightarrow x=\frac{t}{2}\Rightarrow dx=\frac{1}{2}dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=2\Rightarrow t=4\end{matrix}\right.\)
\(\Rightarrow I_2=\int\limits^4_0\frac{t}{2}f'\left(\frac{t}{2}\right).\frac{1}{2}dt=\frac{1}{4}\int\limits^4_0t.f'\left(\frac{t}{2}\right)dt=-2\)
\(\Rightarrow\int\limits^4_0t.f'\left(\frac{t}{2}\right)dt=-8\) hay \(\int\limits^4_0x.f'\left(\frac{x}{2}\right)dx=-8\)
\(\int\limits^2_1\frac{8x+5}{6x^2+7x+2}dx=\int\limits^2_1\frac{8x+5}{6\left(x+\frac{1}{2}\right)\left(x+\frac{2}{3}\right)}dx=\frac{1}{6}\int\limits^2_1(\frac{2}{x+\frac{2}{3}}+\frac{6}{x+\frac{1}{2}})dx\:\)
\(=\frac{1}{6}\left(2ln\left|x+\frac{1}{2}\right|+6ln\left|x+\frac{2}{3}\right|\right)\)\(|^2_1\)
=\(\frac{1}{3}ln\left(\left|x+\frac{1}{2}\right|\right)+ln\left(\left|x+\frac{2}{3}\right|\right)\)\(|^2_1\)
= \(\frac{1}{3}ln\frac{5}{2}+ln\frac{8}{3}-\frac{1}{3}ln\frac{3}{2}-ln\frac{5}{3}=\frac{1}{3}ln5-\frac{1}{3}ln3+ln8-ln3=3ln2-\frac{4}{3}ln3+\frac{1}{3}ln5\)
\(\Rightarrow\)a=3,b=\(\frac{-4}{3}\),c=\(\frac{1}{3}\)
P=2