Đang học Lý mà thấy bài nguyên hàm hay hay nên nhảy vô luôn :b

\(I_1=\int\limits^1_0xf\left(x\right)dx\)

\(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)

\(\Rightarrow\int xf\left(x\right)dx=\dfrac{1}{2}x^2f\left(x\right)-\dfrac{1}{2}\int x^2f'\left(x\right)dx\)

\(\Rightarrow\int\limits^1_0xf\left(x\right)dx=\dfrac{1}{2}x^2|^1_0-\dfrac{1}{2}\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{2}\int\limits^1_0\left[f'\left(x\right)\right]^2dx=\dfrac{3}{10}\Rightarrow\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{3}{5}\)

Đoạn này hơi rối xíu, ông để ý kỹ nhé, nhận thấy ta có 2 dữ kiện đã biết, là: \(\int\limits^1_0\left[f'\left(x\right)\right]^2dx=\dfrac{9}{5}and\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{3}{5}\) có gì đó liên quan đến hằng đẳng thức, nên ta sẽ sử dụng luôn

\(\int\limits^1_0\left[f'\left(x\right)+tx^2\right]^2dx=0\)

\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+2t\int\limits^1_0x^2f'\left(x\right)dx+t^2\int\limits^1_0x^4dx=0\)

\(\Leftrightarrow\dfrac{9}{5}+\dfrac{6}{5}t+\dfrac{1}{5}t^2=0\)  \(\left(\int\limits^1_0x^4dx=\dfrac{1}{5}x^5|^1_0=\dfrac{1}{5}\right)\)\(\)\(\Leftrightarrow t=-3\Rightarrow\int\limits^1_0\left[f'\left(x\right)-3x^2\right]^2dx=0\)

\(\Leftrightarrow f'\left(x\right)=3x^2\Leftrightarrow f\left(x\right)=x^3+C\)

\(\Rightarrow\int\limits^1_0f\left(x\right)dx=\int\limits^1_0x^3dx=\dfrac{1}{4}x^4|^1_0=\dfrac{1}{4}\)

P/s: Có gì ko hiểu hỏi mình nhé !

\(f'\left(x\right)=f'\left(1-x\right)\Rightarrow\int f'\left(x\right)dx=\int f'\left(1-x\right)dx\)

\(\Rightarrow f\left(x\right)=-f\left(1-x\right)+C\Rightarrow f\left(x\right)+f\left(1-x\right)=C\)

Thay \(x=0\Rightarrow f\left(0\right)+f\left(1\right)=C\Rightarrow C=42\)

\(\Rightarrow\int\limits^1_0\left[f\left(x\right)+f\left(1-x\right)\right]dx=\int\limits^1_042dx=42\)

Xét \(I=\int\limits^1_0f\left(1-x\right)dx\)

Đặt \(1-x=u\Rightarrow dx=-du;\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^0_1f\left(u\right).\left(-du\right)=\int\limits^1_0f\left(u\right).du=\int\limits^1_0f\left(x\right)dx\)

\(\Rightarrow2\int\limits^1_0f\left(x\right)dx=42\Rightarrow\int\limits^1_0f\left(x\right)dx=21\)

Lâu lắm mới thấy bà :3

\(f\left(0\right)=-1\Rightarrow f'\left(0\right)+2=0\Leftrightarrow f'\left(0\right)=-2\)

\(\int\limits^1_0f\left(x\right)dx=\int\limits^1_0\dfrac{f'\left(x\right)-x.e^{3x}}{2}dx=\dfrac{1}{2}\int\limits^1_0f'\left(x\right)dx-\dfrac{1}{2}\int\limits^1_0x.e^{3x}dx=\dfrac{1}{2}f\left(x\right)|^1_0-\dfrac{1}{2}\int\limits^1_0xe^{3x}dx\)

\(I_1=\int xe^{3x}dx\)

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

\(\Rightarrow I_1=\dfrac{1}{3}xe^{3x}-\dfrac{1}{3}\int e^{3x}dx=\dfrac{1}{3}xe^{3x}-\dfrac{1}{9}e^{3x}\)

\(\Rightarrow I=\dfrac{1}{2}f\left(1\right)-\dfrac{1}{2}f\left(0\right)-\dfrac{1}{2}\left(\dfrac{1}{3}xe^{3x}-\dfrac{1}{9}e^{3x}\right)|^1_0\)

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 \(x=0\Rightarrow1.f\left(0\right)=-1+C\Rightarrow C=0\)

\(\Rightarrow e^{-2x}.f\left(x\right)=\left(x-1\right)e^x\Rightarrow f\left(x\right)=\left(x-1\right)e^{3x}\)

\(\Rightarrow I=\int\limits^1_0\left(x-1\right)e^{3x}dx=...\)

Chọn D.

Xét  I = ∫ 0 1 f ' x d x   Đặt  t = x → t 2 = x → 2 t d t = d x

Đổi cận   x = 0 → t = 0 x = 1 → t = 1 . Khi đó  I = 2 ∫ 0 1 t f ' ( t ) d t = 2 A

Tính   A = ∫ 0 1 t f ' ( t ) d t . Đặt  u = t d v = f ' t d t → d u = d t v = f t

Khi đó 

\(f'\left(x\right)=-4x^3\left(f\left(x\right)\right)^2\Leftrightarrow-\dfrac{f'\left(x\right)}{\left(f\left(x\right)\right)^2}=4x^3\)

Lấy nguyên hàm hai vế

\(\int-\dfrac{f'\left(x\right)}{\left(f\left(x\right)\right)^2}dx=\int4x^3dx\)

\(\Leftrightarrow\dfrac{1}{f\left(x\right)}=x^4+c\)

Thay x=0 vào tìm được c=1 \(\Rightarrow f\left(x\right)=\dfrac{1}{x^4+1}\)

\(I=\int\limits^1_0\dfrac{x^3}{x^4+1}dx=\dfrac{1}{4}\int\limits^1_0\dfrac{\left(x^4+1\right)'}{x^4+1}dx=\dfrac{ln2}{4}\)

Chọn D

\(3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]dx\le2\int\limits^1_0\sqrt{f'\left(x\right)}f\left(x\right)dx\) (1)

Ta lại có:

\(3f'\left(x\right).f^2\left(x\right)+\frac{1}{3}\ge2\sqrt{f'\left(x\right)}.f\left(x\right)\)

\(\Rightarrow3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]\ge2\int\limits^1_0\sqrt{f'\left(x\right)}.f\left(x\right)dx\) (2)

Từ (1); (2) \(\Rightarrow3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]dx=2\int\limits^1_0\sqrt{f'\left(x\right)}.f\left(x\right)dx\)

Dấu "=" xảy ra khi và chỉ khi:

\(3f'\left(x\right).f^2\left(x\right)=\frac{1}{3}\Rightarrow3\int f'\left(x\right).f^2\left(x\right)dx=\int\frac{1}{3}dx\)

\(\Rightarrow f^3\left(x\right)=\frac{x}{3}+C\)

Thay \(x=0\Rightarrow f^3\left(0\right)=C\Rightarrow C=1\)

\(\Rightarrow f^3\left(x\right)=\frac{x}{3}+1\Rightarrow\int\limits^1_0f^3\left(x\right)dx=\int\limits^1_0\left(\frac{x}{3}+1\right)dx=\frac{7}{6}\)