Cho f(x) = \(\dfrac{1}{2e^x +3}\)thỏa mãn F(0)=10.Tìm F(x)
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\(F\left(x\right)=\int\left(e^x.ln\left(ax\right)+\dfrac{e^x}{x}\right)dx=\int e^xln\left(ax\right)dx+\int\dfrac{e^x}{x}dx=\int e^xlnxdx+\int\dfrac{e^x}{x}dx+\int e^x.lna.dx\)
Xét \(I=\int e^xlnxdx\)
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=lnx.e^x-\int\dfrac{e^x}{x}dx\)
\(\Rightarrow F\left(x\right)=e^x.lnx+e^x.lna+C\)
\(F\left(\dfrac{1}{a}\right)=e^{\dfrac{1}{a}}ln\left(\dfrac{1}{a}\right)+e^{\dfrac{1}{a}}.lna+C=0\Rightarrow C=0\)
\(F\left(2020\right)=e^{2020}ln\left(2020\right)+e^{2020}.lna=e^{2020}\)
\(\Rightarrow ln\left(2020a\right)=1\Rightarrow a=\dfrac{e}{2020}\)
Chọn B
Ta có:
Mặt khác f(0) = 10 => 5 + C = 10 <=> C = 5.
Vậy f(x) = 3x + 5cos x + 5.
Đáp án A
Ta có: f x = ∫ f ' x d x = x 2 2 - cos x + 2 . Do f 0 = 1 ⇒ C = 2 .
\(u=e^x\Rightarrow du=e^xdx\Rightarrow dx=\dfrac{du}{e^x}\)
\(\Rightarrow\int f\left(x\right)dx=\int\dfrac{du}{2u^2+3u}\)
\(\dfrac{1}{2u^2+3u}=\dfrac{A}{u}-\dfrac{B}{2u+3}=\dfrac{A\left(2u+3\right)-Bu}{2u^2+3u}=\dfrac{\left(2A-B\right)u+3A}{2u^2+3u}\)
\(\Rightarrow\left(2A-B\right)u+3A=1\Rightarrow\left\{{}\begin{matrix}2A-B=0\\3A=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=\dfrac{1}{3}\\B=\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\int\dfrac{du}{2u^2+3u}=\dfrac{1}{3}\int\left(\dfrac{1}{u}-\dfrac{2}{2u+3}\right)du=\dfrac{1}{3}\left[lnu-ln\left(2u+3\right)\right]+C\)
\(\Rightarrow F\left(x\right)=\dfrac{1}{3}\left[ln\left(e^x\right)-ln\left(2e^x+3\right)\right]+C=\dfrac{1}{3}\left[x-ln\left(2e^x+3\right)\right]+C\)
\(F\left(0\right)=10\Rightarrow\dfrac{1}{3}\left[x-ln\left(2e^x+3\right)\right]+C=10\Rightarrow C=\dfrac{ln5}{3}+10\)
\(\Rightarrow F\left(x\right)=\dfrac{1}{3}\left[x-ln\left(2e^x+3\right)\right]+\dfrac{ln5}{3}+10\)