48 câu Trắc nghiệm Nguyên hàm có đáp án (P2) (Vận dụng- Vận dụng cao)

Đáp án C

 $x^2+2x-3$ là một nguyên hàm của hàm số $f\left(x\right){.5}^{\frac{x}{2}}$

$\begin{array}{l}\Rightarrow {\left(x^2+2x-3\right)}^{\text{'}}=f\left(x\right){.5}^{\frac{x}{2}}\\ \iff 2x+2=f\left(x\right){.5}^{\frac{x}{2}}\iff f\left(x\right)=\frac{2x+2}{5^{\frac{x}{2}}}\end{array}$

$\begin{array}{l}\Rightarrow f^{\text{'}}\left(x\right)={\left(\frac{2x+2}{5^{\frac{x}{2}}}\right)}^{\text{'}}=\frac{{2.5}^{\frac{x}{2}}-\left(2x+2\right)\left(\frac{1}{2}{.5}^{\frac{x}{2}}.\ln 5\right)}{{\left(5^{\frac{x}{2}}\right)}^2}=\frac{2-\left(x+1\right)\ln 5}{5^{\frac{x}{2}}}\\ \Rightarrow f^{\text{'}}\left(x\right){.5}^{\frac{x}{2}}=2-\left(x+1\right)\ln 5\end{array}$

$\Rightarrow \int f^{\text{'}}\left(x\right){.5}^{\frac{x}{2}}dx=\int \left[2-\left(x+1\right)\ln 5\right]dx=2x-\left(\frac{x^2}{2}+x\right)\ln 5+C.$

Đáp án D

Tacó:

$f^{\text{'}}\left(x\right)\sin x=f\left(x\right)cos\text{x\hspace{0.17em}}+2{\sin }^{2}x.cos3\text{x\hspace{0.17em}};\forall x\in \left(0;\pi \right)$

$\iff \frac{f^{\text{'}}\left(x\right)\sin x-f\left(x\right)cos\text{x\hspace{0.17em}}}{{\sin }^{2}x}=2cos3\text{x\hspace{0.17em}}$

$\Rightarrow {\left(\frac{f\left(x\right)}{\sin x}\right)}^{\text{'}}=2cos3\text{x}\Rightarrow \frac{f\left(x\right)}{\sin x}=\int 2cos3\text{x\hspace{0.17em}dx}$

$\Rightarrow \frac{f\left(x\right)}{\sin x}=\frac{2}{3}\sin 3\text{x\hspace{0.17em}}+C_1$

$f\left(\frac{\pi }{4}\right)=\frac{1}{3}\Rightarrow C_1=0\Rightarrow f\left(x\right)=\frac{2}{3}sinx.\sin 3x$

$\Rightarrow \int f\left(x\right)\text{d}x=\int \frac{2}{3}sinx.\sin 3x\text{\hspace{0.17em}d}x\text{\hspace{0.17em}=\hspace{0.17em}}\frac{1}{3}\int \left(cos2x-\cos 4x\right)\text{dx\hspace{0.17em}}$

$=\frac{1}{12}(2\sin 2\text{x\hspace{0.17em}\hspace{0.17em}}-\sin 4\text{x) \hspace{0.17em}+\hspace{0.17em}C}$

Đáp án C

Ta có $\int f\left(x\right)e^{x}dx=x+\sin x+C\Rightarrow f\left(x\right).e^x=1+\cos x\left(1\right)$

Lấy đạo hàm hai vế của ta được

$f^{\text{'}}\left(x\right).e^x+f\left(x\right).e^x=-\sin x\iff f^{\text{'}}\left(x\right).e^x=-\sin x-\cos x-1$

Vậy ta có $\int f^{\text{'}}\left(x\right).e^{x}dx=-\int \left(\sin x+\cos x+1\right)dx=\cos x-\sin x-x+C$

Đáp án A

Ta có: $f\left(x\right)=\left\{\begin{array}{l}{e}^x+C_1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi\hspace{0.17em}\hspace{0.17em}}x\ge 0\\ -e^{-x}+C_2\text{\hspace{0.17em}\hspace{0.17em}khi\hspace{0.17em}\hspace{0.17em}}x<0\end{array}\right.$ với  $C_1$ và $C_2$ là các hằng số thực.

$f\left(4\right)=e\Rightarrow C_1=e-e^4$

$\underset{x\to 0^{+}}{\lim }f\left(x\right)=1+e-e^4$

$\underset{x\to 0^{-}}{\lim }f\left(x\right)=-1+C_2$

Do hàm số có đạo hàm $\forall x\in ℝ\Rightarrow$ Hàm số liên tục trên $ℝ$.

$\Rightarrow -1+C_2=1+e-e^4=f\left(0\right)\Rightarrow C_2=2+e-e^4$

Vậy $f\left(x\right)=\left\{\begin{array}{l}{e}^x+e-e^4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi\hspace{0.17em}\hspace{0.17em}}x\ge 0\\ -e^{-x}+2+e-e^4\text{\hspace{0.17em}\hspace{0.17em}khi\hspace{0.17em}\hspace{0.17em}}x<0\end{array}\right.$

$S=f\left(-\ln 3\right)+f\left(\ln 3\right)+f\left(-\ln 2\right)+f\left(\ln 2\right)+200=\frac{49}{6}+4e-4e^4+200\approx 0,6$

Đáp án D

$I=\int \frac{x^{2002}}{{\left(x+2\right)}^{2005}}\text{\hspace{0.17em}d}x=\int {\left(\frac{x}{x+2}\right)}^{2002}.\frac{1}{x+2}.\frac{1}{{\left(x+2\right)}^2}\text{\hspace{0.17em}\hspace{0.17em}d}x$

Đặt $t=\frac{x}{x+2}\Rightarrow \text{d}t=\frac{2.\text{d}x}{{\left(x+2\right)}^2}$

$I=\int t^{2002}.\frac{1-t}{2}.\frac{\text{d}t}{2}=\frac{1}{4}\int \left(t^{2002}-t^{2003}\right).\text{\hspace{0.17em}d}t=\frac{1}{8012}.t^{2003}-\frac{1}{8016}.t^{2004}+C$

$\Rightarrow I=\frac{1}{8012}.{\left(\frac{x}{x+2}\right)}^{2003}-\frac{1}{8016}.{\left(\frac{x}{x+2}\right)}^{2004}+C$

Đáp án B

Ta có $f\left(x\right).f\text{'}\left(x\right)=\left(2x+1\right)\sqrt{1+f^2\left(x\right)},\forall x\in ℝ$

$\iff \frac{f\left(x\right).f\text{'}\left(x\right)}{\sqrt{1+f^2\left(x\right)}}=2x+1$

$\iff \int \frac{f\left(x\right).f\text{'}\left(x\right)}{\sqrt{1+f^2\left(x\right)}}\text{d}x=\int \left(2x+1\right)\text{d}x$

$\iff \sqrt{1+f^2\left(x\right)}=x^2+x+C$

Vì  $f\left(0\right)=2\sqrt{2}$ nên $C=3$ .

$\Rightarrow f\left(x\right)=\sqrt{{\left(x^2+x+3\right)}^2-1}\Rightarrow f\left(1\right)=2\sqrt{6}$

Đáp án D

Ta có $f\left(x\right)=\int f^{\text{'}}\left(x\right)dx=\int \frac{x}{\sqrt{6-x^2}}dx=-\frac{1}{2}\int \frac{d\left(6-x^2\right)}{\sqrt{6-x^2}}=-\sqrt{6-x^2}+C$

$f\left(\sqrt{2}\right)=-2$ $\Rightarrow C=0$ $\Rightarrow f\left(x\right)=-\sqrt{6-x^2}$

$\underset{0}{\overset{\sqrt{3}}{\int }}f\left(x\right)dx=\underset{0}{\overset{\sqrt{3}}{\int }}-\sqrt{6-x^2}dx=-\frac{3\pi +6}{4}$

Đáp án D

Đặt $\left\{\begin{array}{l}u=x.\ln \left(x+1\right)\\ \text{d}v=\frac{1}{{\left(x+1\right)}^2}\text{d}x\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{d}u=\left(\ln \left(x+1\right)+\frac{x}{x+1}\right)\text{d}x\\ v=-\frac{1}{x+1}\end{array}\right.$

Ta được:

$\int \frac{x\ln \left(x+1\right)}{{\left(x+1\right)}^2}\text{\hspace{0.33em}d}x=-\frac{x\ln \left(x+1\right)}{x+1}+\int \left[\frac{\ln \left(x+1\right)}{x+1}+\frac{x}{{\left(x+1\right)}^2}\right]\text{\hspace{0.33em}d}x$

$=-\frac{x\ln \left(x+1\right)}{x+1}+\int \left[\frac{\ln \left(x+1\right)}{x+1}+\frac{x+1-1}{{\left(x+1\right)}^2}\right]\text{\hspace{0.33em}d}x$

$=-\frac{x\ln \left(x+1\right)}{x+1}+\int \left[\frac{\ln \left(x+1\right)}{x+1}+\frac{1}{x+1}-\frac{1}{{\left(x+1\right)}^2}\right]\text{\hspace{0.33em}d}x$

$=-\frac{x\ln \left(x+1\right)}{x+1}+\frac{1}{2}{\ln }^2\left(x+1\right)+\ln \left(x+1\right)+\frac{1}{x+1}+C$

Đáp án D

Ta có $F\left(x\right)=\int x^2{e}^{\alpha x}\text{d}x=\frac{1}{\alpha }\int x^2\text{d}\left(e^{\alpha x}\right)=\frac{1}{\alpha }\left[x^2{e}^{\alpha x}-\int e^{\alpha x}2x\text{d}x\right]$

$=\frac{1}{\alpha }\left(x^2{e}^{\alpha x}-\frac{2}{\alpha }\int x\text{d}\left(e^{\alpha x}\right)\right)=\frac{1}{\alpha }\left(x^2{e}^{\alpha x}-\frac{2}{\alpha }\left(x{e}^{\alpha x}-\int e^{\alpha x}\text{d}x\right)\right)$

$=\frac{1}{\alpha }\left(x^2{e}^{\alpha x}-\frac{2}{\alpha }x{e}^{\alpha x}+\frac{2}{{\alpha }^2}{e}^{\alpha x}\right)+C$

$F\left(\frac{1}{\alpha }\right)=\frac{1}{\alpha }\left(\frac{1}{{\alpha }^2}e-\frac{2}{{\alpha }^2}e+\frac{2}{{\alpha }^2}e\right)+C=\frac{1}{{\alpha }^3}e+C$

$F\left(0\right)=\frac{1}{\alpha }.\frac{2}{{\alpha }^2}+C$

Theo giả thiết $F\left(\frac{1}{\alpha }\right)-F\left(0\right)=\frac{e-2}{{\alpha }^3}=1\Rightarrow {\alpha }^3=e-2\Rightarrow \alpha =\sqrt[3]{e-2}\Rightarrow 0<\alpha \le 1$