• 2022-06-11
    ∫e^2xsinxdx=?网上有2种答案.1.(1/5)(e^2x)(2sinx-cosx)+c2.∫sinx*e^2xdx=[-cosxe^2x+2sinxe^2x]/3+C
  • I=∫e^2xsinxdx=-∫e^2xdcosx=-e^(2x)cosx+∫cosxde^2x=-e^(2x)cosx+2∫e^2xcosxdx=-e^(2x)cosx+2∫e^2xdsinx=-e^(2x)cosx+2e^(2x)sinx-2∫sinxde^2x=-e^(2x)cosx+2e^(2x)sinx-4∫e^2xsinxdx移项得I=...

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    • 0

      设∫xf(x)dx=arcsinx+C<sub>1</sub>,则∫&#91;1/f(x)&#93;dx=()。 A: (1-x<sup>2</sup>)<sup>3/2</sup>/3+C B: -(1-x<sup>2</sup>)<sup>3/2</sup>/3+C C: (1+x<sup>2</sup>)<sup>3/2</sup>/3+C D: (1+x<sup>2</sup>)<sup>2/3</sup>/3+C

    • 1

      设\(z = \int_ { { x^2}}^y { { e^t}\sin t} dt\),则\({z_{xx}=}\) A: \(2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) B: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} - 2{x^2}\cos {x^2}} \right]\) C: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) D: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\cos {x^2} + 2{x^2}\sin {x^2}} \right]\)

    • 2

      积分[img=136x52]1803d6afd4e6f95.png[/img]的计算程序和结果是 A: clearsyms xy=1/x^2/sqrt(x^2-1)int(y,x,-2,-1)3^(1/2)/2 B: clearsyms xint(1/x^2/sqrt(x^2-1),x,-2,-1)3^(1/2)/2 C: clearsyms xint(1/x/sqrt(x^2-1),x,-2,-1)-pi/3 D: clearsyms xint(1/x/sqrt(x^2-1),x,-2,-1)3^(1/2)/2 E: clearsyms xint(1/x^2*sqrt(x^2-1),x,-2,-1)log(3^(1/2) + 2) - 3^(1/2)/2

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      利用 Mean Value Theorem 求 \(lim_{x\rightarrow 0+}\frac{2^{x}-2^{sinx}}{x-sinx}\) 的解 (Mean Value Theorem을 이용하여 \(lim_{x\rightarrow 0+}\frac{2^{x}-2^{sinx}}{x-sinx}\)의 값을 구하여라). A: \(ln\, 2\) B: \(2ln\, 2\) C: \(e^{x}\) D: \(e^{2x}\) E: \(2e\)

    • 4

      数学式 A: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sinx^2-Cos2x)) B: (Exp(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) C: (Exp(2*x)*Ln(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) D: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x)^2-Cos(x)^2))