Tính f(x)=\(\int e^2dx\), trong đó e là hằng số và e\(\approx\)2,718
A. f(x)= \(\dfrac{e^2x^2}{2}+C\)
B. f(x) =\(\dfrac{e^3}{3}+C\)
C. f(x) = e\(^2\)x+C
D. f(x) = 2ex + C
Tính f(x)=\(\int e^2dx\), trong đó e là hằng số và e\(\approx\)2,718
A. f(x)= \(\dfrac{e^2x^2}{2}+C\)
B. f(x) =\(\dfrac{e^3}{3}+C\)
C. f(x) = e\(^2\)x+C
D. f(x) = 2ex + C
Cho hàm số f(x) liên tục trên R Biết cận 0->pi/2 sin2x f(cos^2(x)) dx =1 Khi đó cân 0->1[2f(1-x) -3x^2+5]dx=?
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 \(cos^2x=1-u\Rightarrow-2sinx.cosxdx=-du\) \(\Rightarrow sin2xdx=du\)
\(\left\{{}\begin{matrix}x=0\Rightarrow u=0\\x=\dfrac{\pi}{2}\Rightarrow u=1\end{matrix}\right.\) \(\Rightarrow I=\int\limits^1_0f\left(1-u\right)du=\int\limits^1_0f\left(1-x\right)dx\)
\(\Rightarrow\int\limits^1_0f\left(1-x\right)dx=1\)
\(\Rightarrow\int\limits^1_0\left[2f\left(1-x\right)-3x^2+5\right]dx=2\int\limits^1_0f\left(1-x\right)dx-\int\limits^1_0\left(3x^2-5\right)dx\)
\(=2.1-\left(-4\right)=6\)
tìm nguyên hàm của hàm số y=x^(2)-3cosx+(1)/(x)
1.(4x+1)e^(x) giúp ạ
\(\int\left(4x+1\right)e^xdx\)
Đặt \(\left\{{}\begin{matrix}u=4x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=4dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=\left(4x+1\right)e^x-\int4e^xdx=\left(4x+1\right)e^x-4e^x+C\)
\(=\left(4x-3\right)e^x+C\)
Tính:
\(A=\int\dfrac{cosx}{\left(x+1\right)^2}dx\)
1) Tinh nguyen ham
a) A = \(\int\dfrac{x}{\sqrt{x+2}}.dx\) b) B = \(\int\dfrac{sinx+cosx}{\sqrt[3]{1-sin2x}}.dx\)
\(A=\int \frac{x}{\sqrt{x+2}}dx \\ = \int \frac{x+2-2}{\sqrt{x+2}}dx \\ = \int \sqrt{x+2}-2\frac{1}{\sqrt{x+2}}dx \\ = \frac{2}{3}(x+2)^{\frac{3}{2}}-4\sqrt{x+2}+C\)
\(B=\int \frac{sinx+cosx}{\sqrt[3]{1-sin2x}}dx \\ x=\frac{\pi}{4}-u, dx=-du \\ =- \int \frac{sin(\frac{\pi}{4}-u)+cos(\frac{\pi}{4}-u)}{\sqrt[3]{1-sin(\frac{\pi}{2}-2u)}}du \\ = - \int \frac{\frac{1}{\sqrt2}cosu+\frac{1}{\sqrt2}sinu+\frac{1}{\sqrt2}cosu-\frac{1}{\sqrt2}sinu}{\sqrt[3]{1-cos2u}}du \\ = -\int \frac{\frac{2}{\sqrt2}cosu}{\sqrt[3]{1-cos2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{1-cos^2u+sin^2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{2sin^2u}}du \\ v=sinu, dv=cosudu \\ = -\sqrt2 \int \frac{1}{\sqrt[3]{2v^2}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} \int v^{-\frac{2}{3}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} 3v^\frac{1}{3}+C \\ = -\frac{\sqrt2}{\sqrt[3]2} 3\sqrt[3]{sin(\frac{\pi}{4}-x)}+C \)
Mọi người giúp mk giải chi tiết câu này với ạ. Mk cảm ơn
\(I=\int\dfrac{2}{2+5sinxcosx}dx=\int\dfrac{2sec^2x}{2sec^2x+5tanx}dx\\ =\int\dfrac{2sec^2x}{2tan^2x+5tanx+2}dx\)
We substitute :
\(u=tanx,du=sec^2xdx\\ I=\int\dfrac{2}{2u^2+5u+2}du\\ =\int\dfrac{2}{2\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{8}}du\\ =\int\dfrac{1}{\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{16}}du\\ \)
Then,
\(t=u+\dfrac{5}{4}\\I=\int\dfrac{1}{t^2-\dfrac{9}{16}}dt\\ =\int\dfrac{\dfrac{2}{3}}{t-\dfrac{3}{4}}-\dfrac{\dfrac{2}{3}}{t+\dfrac{3}{4}}dt\)
Finally,
\(I=\dfrac{2}{3}ln\left(\left|\dfrac{t-\dfrac{3}{4}}{t+\dfrac{3}{4}}\right|\right)+C=\dfrac{2}{3}ln\left(\left|\dfrac{tanx+\dfrac{1}{2}}{tanx+2}\right|\right)+C\)
Tính \(\int\left(10+\cot^2x\right)dx\)
\(f\left(1\right)-f\left(0\right)=\int_0^1\dfrac{1}{\sqrt{x+1}-\sqrt{x}}dx\)
\(=\int_0^1\left(\sqrt{x+1}+\sqrt{x}\right)dx\)
\(=\left(\dfrac{2}{3}.\left(x+1\right)^{\dfrac{3}{2}}+\dfrac{2}{3}.\left(x\right)^{\dfrac{3}{2}}\right)|_0^1\)
\(=\dfrac{2}{3}\left(2^{\dfrac{3}{2}}+1\right)-\dfrac{2}{3}\)
\(=\dfrac{2}{3}\left(\sqrt{8}+1\right)-\dfrac{2}{3}=\dfrac{4\sqrt{2}}{3}\)
Nguyên hàm của \(\int\limits\dfrac{1}{\left(4+x^2\right)^{\dfrac{3}{2}}}dx\) là:
\(I=\int\dfrac{x^2+4-x^2}{4\sqrt{\left(x^2+4\right)^3}}dx=\dfrac{1}{4}\int\dfrac{1}{\sqrt{x^2+4}}dx-\dfrac{1}{4}\int\dfrac{x^2}{\sqrt{\left(x^2+4\right)^3}}dx\)
Xét \(J=\int\dfrac{1}{\sqrt{x^2+4}}dx\)
Đặt \(\left\{{}\begin{matrix}u=\dfrac{1}{\sqrt{x^2+4}}\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-\dfrac{x}{\sqrt{\left(x^2+4\right)^3}}dx\\v=x\end{matrix}\right.\)
\(\Rightarrow J=\dfrac{x}{\sqrt{x^2+4}}+\int\dfrac{x^2}{\sqrt{\left(x^2+4\right)^3}}dx\)
\(\Rightarrow I=\dfrac{x}{4\sqrt{x^2+4}}+\dfrac{1}{4}\int\dfrac{x^2}{\sqrt{\left(x^2+4\right)^3}}dx-\dfrac{1}{4}\int\dfrac{x^2}{\sqrt{\left(x^2+4\right)^3}}dx=\dfrac{x}{4\sqrt{x^2+4}}+C\)