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Lời giải:
Bài 1:
Ta nhớ công thức \(\sin^2x=\frac{1-\cos 2x}{2}\). Áp dụng vào bài toán:
\(F(x)=8\int \sin^2\left(x+\frac{\pi}{12}\right)dx=4\int \left [1-\cos \left(2x+\frac{\pi}{6}\right)\right]dx\)
\(\Leftrightarrow F(x)=4\int dx-4\int \cos \left(2x+\frac{\pi}{6}\right)dx=4x-2\int \cos (2x+\frac{\pi}{6})d(2x+\frac{\pi}{6})\)
\(\Leftrightarrow F(x)=4x-2\sin (2x+\frac{\pi}{6})+c\)
Giải thích 1 chút: \(d(2x+\frac{\pi}{6})=(2x+\frac{\pi}{6})'dx=2dx\)
Vì \(F(0)=8\Rightarrow -1+c=8\Rightarrow c=9\)
\(\Rightarrow F(x)=4x-2\sin (2x+\frac{\pi}{6})+9\)
Câu 2:
Áp dụng nguyên hàm từng phần như bài bạn đã đăng:
\(\Rightarrow F(x)=-xe^{-x}-e^{-x}+c\)
Vì \(F(0)=1\Rightarrow -1+c=1\Rightarrow c=2\)
\(\Rightarrow F(x)=-e^{-x}(x+1)+2\), tức B là đáp án đúng
Cho hàm số y=f(x)y=f(x) có đạo hàm và liên tục trên [0;π2][0;π2]thoả mãn f(x)=f′(x)−2cosxf(x)=f′(x)−2cosx. Biết f(π2)=1f(π2)=1, tính giá trị f(π3)f(π3)
A. √3+1/2 B. √3−1/2 C. 1−√3/2 D. 0
\(2x.f'\left(x\right)-f\left(x\right)=x^2\sqrt{x}.cosx\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}.f'\left(x\right)-\dfrac{1}{2x\sqrt{x}}f\left(x\right)=x.cosx\)
\(\Leftrightarrow\left[\dfrac{f\left(x\right)}{\sqrt{x}}\right]'=x.cosx\)
Lấy nguyên hàm 2 vế:
\(\int\left[\dfrac{f\left(x\right)}{\sqrt{x}}\right]'dx=\int x.cosxdx\)
\(\Rightarrow\dfrac{f\left(x\right)}{\sqrt{x}}=x.sinx+cosx+C\)
\(\Rightarrow f\left(x\right)=x\sqrt{x}.sinx+\sqrt{x}.cosx+C.\sqrt{x}\)
Thay \(x=4\pi\)
\(\Rightarrow0=4\pi.\sqrt{4\pi}.sin\left(4\pi\right)+\sqrt{4\pi}.cos\left(4\pi\right)+C.\sqrt{4\pi}\)
\(\Rightarrow C=-1\)
\(\Rightarrow f\left(x\right)=x\sqrt{x}.sinx+\sqrt{x}.cosx-\sqrt{x}\)
a: \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
=>\(y'=\dfrac{1}{3}\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}\cdot\left(2x^2-x+1\right)'\)
\(=\dfrac{1}{3}\cdot\left(4x-1\right)\left(2x^2-x+1\right)^{-\dfrac{2}{3}}\)
b: \(y=\left(3x+1\right)^{\Omega}\)
=>\(y'=\Omega\cdot\left(3x+1\right)'\cdot\left(3x+1\right)^{\Omega-1}\)
=>\(y'=3\Omega\left(3x+1\right)^{\Omega-1}\)
c: \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
=>\(y'=\dfrac{\left(\dfrac{1}{x-1}\right)'}{3\cdot\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(=\dfrac{\dfrac{1'\left(x-1\right)-\left(x-1\right)'\cdot1}{\left(x-1\right)^2}}{\dfrac{3}{\sqrt[3]{\left(x-1\right)^2}}}\)
\(=\dfrac{-x}{\left(x-1\right)^2}\cdot\dfrac{\sqrt[3]{\left(x-1\right)^2}}{3}\)
\(=\dfrac{-x}{\sqrt[3]{\left(x-1\right)^4}\cdot3}\)
d: \(y=log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Leftrightarrow y'=\dfrac{\left(\dfrac{x+1}{x-1}\right)'}{\dfrac{x+1}{x-1}\cdot ln3}\)
\(\Leftrightarrow y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}:\dfrac{ln3\left(x+1\right)}{x-1}\)
\(\Leftrightarrow y'=\dfrac{x-1-x-1}{\left(x-1\right)^2}\cdot\dfrac{x-1}{ln3\cdot\left(x+1\right)}\)
\(\Leftrightarrow y'=\dfrac{-2}{\left(x-1\right)\cdot\left(x+1\right)\cdot ln3}\)
e: \(y=3^{x^2}\)
=>\(y'=\left(x^2\right)'\cdot ln3\cdot3^{x^2}=2x\cdot ln3\cdot3^{x^2}\)
f: \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
=>\(y'=\left(x^2-1\right)'\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}=2x\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}\)
h: \(y=\left(x+1\right)\cdot e^{cosx}\)
=>\(y'=\left(x+1\right)'\cdot e^{cosx}+\left(x+1\right)\cdot\left(e^{cosx}\right)'\)
=>\(y'=e^{cosx}+\left(x+1\right)\cdot\left(cosx\right)'\cdot e^u\)
\(=e^{cosx}+\left(x+1\right)\cdot\left(-sinx\right)\cdot e^u\)
a) \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}.\left(4x-1\right)\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{-\dfrac{2}{3}}.\left(4x-1\right)\)
b) \(y=\left(3x+1\right)^{\pi}\)
\(\Rightarrow y'=\pi.\left(3x+1\right)^{\pi-1}.3=3\pi.\left(3x+1\right)^{\pi-1}\)
c) \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
\(\Rightarrow y'=\dfrac{\left(x-1\right)^{-1-1}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^{3-1}}}=\dfrac{\left(x-1\right)^{-2}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}=\dfrac{1}{3.\sqrt[]{x-1}.\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(\Rightarrow y'=\dfrac{1}{3\left(x-1\right)^{\dfrac{1}{2}}.\left(x-1\right)^{\dfrac{2}{3}}}=\dfrac{1}{3\left(x-1\right)^{\dfrac{7}{6}}}=\dfrac{1}{3\sqrt[6]{\left(x-1\right)^7}}\)
d) \(y=\log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Rightarrow y'=\dfrac{\dfrac{1-\left(-1\right)}{\left(x-1\right)^2}}{\dfrac{x+1}{x-1}.\ln3}=\dfrac{2}{\left(x+1\right)\left(x-1\right).\ln3}\)
e) \(y=3^{x^2}\)
\(\Rightarrow y'=3^{x^2}.ln3.2x=2x.3^{x^2}.ln3\)
f) \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
\(\Rightarrow y'=\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}.2x=2x.\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}\)
Các bài còn lại bạn tự làm nhé!
đặt :
\(F\left(x\right)=\int_0^{x^2}f\left(t\right)dt=xsin\left(\pi x\right)\Leftrightarrow F\left(x^2\right)-F\left(0\right)=xsin\)
\(\left(\pi x\right)\Leftrightarrow F\left(x^2\right)=F\left(0\right)+xsin\left(\pi x\right)\)
lấy đạo hàm \(2\) vế , ta có :
\(\left(F\left(0\right)\right)'=sin\left(\pi x\right)+\pi xcos\left(\pi x\right)+\left(F\left(0\right)\right)'\)
\(\Leftrightarrow2xf\left(x^2\right)=sin\left(\pi x\right)+\pi xcos\left(\pi x\right)\)
thay \(x=2\) , ta có :
\(2.2.f\left(4\right)=sin\left(2\pi\right)+2\pi cos\left(2\pi\right)\Leftrightarrow4f\left(4\right)=2\pi\Leftrightarrow f\left(4\right)=\dfrac{\pi}{2}\)
Đề là cho \(\int\limits^{\dfrac{\pi}{2}}_0sin2x.f\left(cos^2x\right)dx=1\)
Tính \(\int\limits^1_0\left[2f\left(1-x\right)-3x^2+5\right]dx\)
Đúng ko nhỉ?
Xét \(\int\limits^{\dfrac{\pi}{2}}_0sin2x.f\left(cos^2x\right)dx\)
Đặt \(cos^2x=1-u\Rightarrow-2sinx.cosxdx=-du\) \(\Rightarrow sin2xdx=du\)
\(\left\{{}\begin{matrix}x=0\Rightarrow u=0\\x=\dfrac{\pi}{2}\Rightarrow u=1\end{matrix}\right.\) \(\Rightarrow I=\int\limits^1_0f\left(1-u\right)du=\int\limits^1_0f\left(1-x\right)dx\)
\(\Rightarrow\int\limits^1_0f\left(1-x\right)dx=1\)
\(\Rightarrow\int\limits^1_0\left[2f\left(1-x\right)-3x^2+5\right]dx=2\int\limits^1_0f\left(1-x\right)dx-\int\limits^1_0\left(3x^2-5\right)dx\)
\(=2.1-\left(-4\right)=6\)
Lời giải:
Ta có \(F(x)=\int \sin xe^{\cos x}dx=-\int e^{\cos x}d(\cos x)\)
\(\Leftrightarrow F(x)=-e^{\cos x}+c\)
Mà \(F(0)=e+c=e\Rightarrow c=0\)
\(\Rightarrow F(\pi)=-e^{\cos \pi}=\frac{-1}{e}\). Đáp án B