200 Bài tập nguyên hàm tích phân cơ bản, nâng cao (P1)

Đáp án D.

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: \int \mathrm{f}(\mathrm{x}-1)\mathrm{dx}=\frac{\mathrm{x}-1}{\mathrm{x}+1}+\mathrm{C} \\ \iff \mathrm{f}(\mathrm{x}-1)=\frac{2}{{(\mathrm{x}+1)}^2}=\frac{2}{{\left[(\mathrm{x}-1)+2\right]}^2} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{2}{{(\mathrm{x}+2)}^2} \\ \Rightarrow \mathrm{f}(\mathrm{x}+1)=\frac{2}{{(\mathrm{x}+1+2)}^2}=\frac{2}{{(\mathrm{x}+3)}^2} \\ \Rightarrow \int \mathrm{f}(\mathrm{x}+1)\mathrm{dx}=\int \frac{2}{{(\mathrm{x}+3)}^2}\mathrm{dx}=\frac{-2}{\mathrm{x}+3}+\mathrm{C} \\ \Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{-2}{\mathrm{x}+3} \end{gathered}$$

Đáp án B.

$$\begin{gathered} \frac{\sin 4x}{{\sin }^{4}x+{\cos }^{4}x}=\frac{2\sin 2x\cos 2x}{1-2{\sin }^{2}x{\cos }^{2}x}=\frac{2\sin 2x\cos 2x}{1-2.{\displaystyle \frac{1}{4}}{\sin }^{2}2x}=\frac{4\sin 2x\cos 2x}{2-{\sin }^{2}2x} \\ \Rightarrow \int \frac{\sin 4xdx}{{\sin }^{4}x+{\cos }^{4}x}=\int \frac{4\sin 2x\cos 2xdx}{2-{\sin }^{2}2x}=I \\ Đặt 2-{\sin }^{2}2x=t \iff -4\sin 2x\cos 2xdx=dt \\ \Rightarrow I=\int \frac{-dt}{t}=-\ln \left|t\right|=-\ln \left| 2-{\sin }^{2}2x\right|=\ln \left|\frac{1}{ 2-{\sin }^{2}2x}\right| \\ =\ln \left|\frac{1}{{\sin }^{4}x+{\cos }^{4}x}\right|=\ln \left(\frac{1}{{\sin }^{4}x+{\cos }^{4}x}\right) \left(do \frac{1}{{\sin }^{4}x+{\cos }^{4}x}>0\right) \end{gathered}$$

Đáp án A.

$$\begin{gathered} \mathrm{Đặt} {\mathrm{x}}^2=\mathrm{t}\iff \mathrm{xdx}=\frac{\mathrm{dt}}{2} \\ \mathrm{Đổi} \mathrm{cận} \mathrm{x}=0\Rightarrow \mathrm{t}=0; \mathrm{x}=1\Rightarrow \mathrm{t}=1 \\ \Rightarrow \mathrm{I}={\int }_0^1\mathrm{f}\left(\mathrm{t}\right).\frac{1}{2}.d\mathrm{t}=\frac{1}{2}{\int }_0^1\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x}=\frac{1}{2}.2=1 \end{gathered}$$

Đáp án C.

Đáp án D.

$$\begin{gathered} \mathrm{I}={\int }_{-2}^2\left[\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}(-\mathrm{x})\right]d\mathrm{x}={\int }_{-2}^2\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x}-{\int }_{-2}^2\mathrm{f}(-\mathrm{x})d\mathrm{x}=\frac{\mathrm{\pi }}{2}-\mathrm{K} \\ \mathrm{Đặt} -\mathrm{x}=\mathrm{t}\iff \mathrm{dx}=-\mathrm{dt} \\ \mathrm{Đổi} \mathrm{cận} \mathrm{x}=2\Rightarrow \mathrm{t}=-2; \mathrm{x}=-2\Rightarrow \mathrm{t}=2 \\ \Rightarrow \mathrm{K}={\int }_2^{-2}-\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}={\int }_{-2}^2\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}={\int }_{-2}^2\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x}=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \mathrm{I}=0 \end{gathered}$$

Đáp án C.

$$\begin{gathered} x^2+{(y-2)}^2=1\iff {\left(y-2\right)}^2=1-x^2 \Rightarrow ĐK: 1-x^2>0\iff -1<x<1 \\ \iff \left\{\begin{array}{l}y=2+\sqrt{1-x^2}\\ y=2-\sqrt{1-x^2}\end{array}\right. \\ \Rightarrow V=\mathrm{\pi }{\int }_{-1}^1\left[{\left(2+\sqrt{1-{\mathrm{x}}^2}\right)}^2-{\left(2-\sqrt{1-{\mathrm{x}}^2}\right)}^2\right]dx \\ =\mathrm{\pi }{\int }_{-1}^{1}8\sqrt{1-{\mathrm{x}}^2}d\mathrm{x}=8\mathrm{\pi }{\int }_{-1}^1\sqrt{1-{\mathrm{x}}^2}d\mathrm{x} \end{gathered}$$

Đáp án D.

$$\begin{gathered} Xét PT: x^3-4x^2+x+6=0\iff \left\{\begin{array}{l}x=-1\\ x=3\\ x=2\end{array}\right. \\ Xét dấu biểu thức f\left(x\right)=x^3-4x^2+x+6 ta thấy \\ f\left(x\right)>0\iff -1<x<2 và \\ f\left(x\right)<0\iff 2<x<3 \\ \Rightarrow D={\int }_{-1}^{2}f\left(x\right)dx-{\int }_2^{3}f\left(x\right)dx \end{gathered}$$

Đáp án D.

Đáp án C.

$$\begin{gathered} \int \mathrm{cosx}.{\sin }^6\mathrm{xdx}=\int {\sin }^6\mathrm{x}.\mathrm{d}\left(\mathrm{sinx}\right)=\frac{1}{7}.{\sin }^7\mathrm{x}+\mathrm{C} \\ \Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{1}{7}.{\sin }^7\mathrm{x} \end{gathered}$$

Đáp án C.

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: \left|1-2\mathrm{x}\right|=\left\{\begin{array}{l}1-2\mathrm{x} \mathrm{khi} \mathrm{x}\le \frac{1}{2}\\ 2\mathrm{x}-1 \mathrm{khi} \mathrm{x}>\frac{1}{2}\end{array}\right. \\ \Rightarrow \mathrm{I}={\int }_0^2\left|1-2\mathrm{x}\right|d\mathrm{x}={\int }_0^{\frac{1}{2}}(1-2\mathrm{x})d\mathrm{x}+{\int }_{\frac{1}{2}}^2(2\mathrm{x}-1)d\mathrm{x} \\ =\frac{1}{4}+\frac{9}{4}=\frac{5}{2} \end{gathered}$$

Đáp án A.

Đáp án B.

Diện tích hình phẳng cần tính là phần màu vàng như hình vẽ

$$\begin{gathered} \mathrm{Xét} \mathrm{PT}: {\mathrm{e}}^{\mathrm{x}}={\mathrm{e}}^{-\mathrm{x}}\iff \mathrm{x}=0 \\ \Rightarrow \mathrm{S}={\int }_0^1\left({\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}\right)d\mathrm{x}={\int }_0^1{\mathrm{e}}^{\mathrm{x}}d\mathrm{x}-{\int }_0^1{\mathrm{e}}^{-\mathrm{x}}d\mathrm{x} \\ =\mathrm{e}-1-\left(\frac{-1}{\mathrm{e}}+1\right)=\mathrm{e}+\frac{1}{\mathrm{e}}-2 \end{gathered}$$

Đáp án D.

$Xét PT: x-\mathrm{\pi }=0\iff \mathrm{x}=\mathrm{\pi }$

Đáp án C.

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: \mathrm{I}=\int \frac{\mathrm{x}}{{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx} \\ \mathrm{Đặt} \left\{\begin{array}{l}\mathrm{x}=\mathrm{u}\\ \frac{1}{{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}=\mathrm{dv}\end{array}\iff \left\{\begin{array}{l}\mathrm{dx}=\mathrm{du}\\ \mathrm{v}=-{\mathrm{e}}^{-\mathrm{x}}\end{array}\right.\right. \\ \Rightarrow \mathrm{I}=-\mathrm{x}.{\mathrm{e}}^{-\mathrm{x}}+\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}=-\mathrm{x}.{\mathrm{e}}^{-\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}} \\ \mathrm{I}=-{\mathrm{e}}^{-\mathrm{x}}(\mathrm{x}+1)+\mathrm{C}=-\frac{(\mathrm{x}+1)}{{\mathrm{e}}^{\mathrm{x}}}+\mathrm{C} \end{gathered}$$

Đáp án B.

$$\begin{gathered} \int \cos 2\mathrm{xcosxdx}=\int \frac{1}{2}(\cos 3\mathrm{x}+\mathrm{cosx})\mathrm{dx} \\ =\frac{1}{2}\left(\frac{\sin 3\mathrm{x}}{3}+\mathrm{sinx}\right)+\mathrm{C} \\ =\frac{1}{6}\sin 3\mathrm{x}+\frac{1}{2}\mathrm{sinx}+\mathrm{C} \end{gathered}$$

Đáp án C.

$$\begin{gathered} \mathrm{Ta} \mathrm{có}:\left|\mathrm{x}\right|=\left\{\begin{array}{l}\mathrm{x} \mathrm{nếu} \mathrm{x}\ge 0\\ -\mathrm{x} \mathrm{nếu} \mathrm{x}<0\end{array}\right. \\ \Rightarrow \mathrm{I}={\int }_{-1}^{\mathrm{a}}\left|\mathrm{x}\right|d\mathrm{x}={\int }_{-1}^0-\mathrm{x}d\mathrm{x}+{\int }_0^{\mathrm{a}}\mathrm{x}d\mathrm{x} \\ =\frac{1}{2}+\frac{{\mathrm{a}}^2}{2}=\frac{{\mathrm{a}}^2+1}{2} \end{gathered}$$

Đáp án B.

Đáp án D.

$$\begin{gathered} Xét PT: 6x-3x^2=0\iff \left\{\begin{array}{l}x=0\\ x=2\end{array}\right. \\ \Rightarrow S={\int }_0^2\left|6x-3x^2\right|dx={\int }_0^2\left(6x-3x^2\right)dx=4 \\ (do 6x-3x^2>0 với x\in (0;2\left)\right) \end{gathered}$$

Đáp án B.

$$\begin{gathered} \mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}+\mathrm{cosx})\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{x}}\mathrm{sinxdx}+\int {\mathrm{e}}^{\mathrm{x}}\mathrm{cosxdx} =\mathrm{A}+\mathrm{B} \\ \mathrm{Đặt} \left\{\begin{array}{l}\mathrm{sinx}=\mathrm{u}\\ {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=\mathrm{dv}\end{array}\iff \left\{\begin{array}{l}\mathrm{cosxdx}=\mathrm{du}\\ \mathrm{v}={\mathrm{e}}^{\mathrm{x}}\end{array}\right.\right. \\ \Rightarrow \mathrm{A}={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}-\int {\mathrm{e}}^{\mathrm{x}}\mathrm{cosxdx}={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}-\mathrm{B} \\ \Rightarrow \mathrm{I}={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}-\mathrm{B}+\mathrm{B}={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx} \end{gathered}$$

Đáp án B.

$$\begin{gathered} \mathrm{Ta} \mathrm{có}:\int \cos 3\mathrm{xdx}=\frac{\sin 3\mathrm{x}}{3}+\mathrm{C} \\ \Rightarrow \mathrm{F}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{3}+\mathrm{C}=1\Rightarrow \mathrm{C}=\frac{2}{3} \\ \Rightarrow \mathrm{F}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{\sin \left({\displaystyle \mathrm{\pi }}\right)}{3}+\frac{2}{3}=\frac{2}{3} \end{gathered}$$

Đáp án D.

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: \left|2\mathrm{x}-1\right|=\left\{\begin{array}{l}2\mathrm{x}-1 \mathrm{nếu} \mathrm{x}\ge \frac{1}{2}\\ 1-2\mathrm{x} \mathrm{nếu} \mathrm{x}<\frac{1}{2}\end{array}\right. \\ \Rightarrow {\int }_{-1}^1\left|2\mathrm{x}-1\right|d\mathrm{x}={\int }_{-1}^{\frac{1}{2}}(1-2\mathrm{x})d\mathrm{x}+{\int }_{\frac{1}{2}}^1(2\mathrm{x}-1)d\mathrm{x} \\ =\frac{9}{4}+\frac{1}{4}=\frac{5}{2} \end{gathered}$$

Đáp án D.

Đáp án A.

$$\begin{gathered} \mathrm{Xét} \mathrm{PT}: \frac{{\mathrm{e}}^{\mathrm{x}}-1}{{\mathrm{e}}^{\mathrm{x}}}=0 \iff \mathrm{x}=0 \\ \Rightarrow {\mathrm{S}}_{\mathrm{D}}={\int }_0^1\left(\frac{{\mathrm{e}}^{\mathrm{x}}-1}{{\mathrm{e}}^{\mathrm{x}}}\right)d\mathrm{x}={\int }_0^1\left(1-\frac{1}{{\mathrm{e}}^{\mathrm{x}}}\right)d\mathrm{x} \\ ={\int }_0^{1}d\mathrm{x}-{\int }_0^1\frac{1}{{\mathrm{e}}^{\mathrm{x}}}d\mathrm{x}=1+\frac{1}{\mathrm{e}}-1=\frac{1}{\mathrm{e}} \end{gathered}$$

Đáp án C.

$$\begin{gathered} \mathrm{Gọi} \mathrm{d} \mathrm{là} \mathrm{đường} \mathrm{thẳng} \mathrm{đi} \mathrm{qua} 2 \mathrm{điểm} (2;0) \mathrm{và} (1;1) \mathrm{có} \mathrm{dạng} \mathrm{y}=\mathrm{ax}+\mathrm{b} \\ \Rightarrow \left\{\begin{array}{l}2\mathrm{a}+\mathrm{b}=0\\ \mathrm{a}+\mathrm{b}=1\end{array}\iff \left\{\begin{array}{l}\mathrm{b}=2\\ \mathrm{a}=-1\end{array}\right.\right.\Rightarrow \mathrm{d}: \mathrm{y}=-\mathrm{x}+2 \\ \Rightarrow \mathrm{Thể} \mathrm{tích} \mathrm{của} \mathrm{miền} \mathrm{phẳng} \mathrm{D} \mathrm{giới} \mathrm{hạn} \mathrm{bởi} \left\{\begin{array}{l}\mathrm{y}=\mathrm{x}; \mathrm{y}=-\mathrm{x}+2\\ \mathrm{x}=1; \mathrm{x}=2\end{array}\right.\mathrm{khi} \mathrm{quay} \mathrm{D} \mathrm{quanh} \mathrm{trục} \mathrm{Ox} \mathrm{là} \\ \mathrm{V}=\mathrm{\pi }{\int }_1^2\left[{\mathrm{x}}^2-{(2-\mathrm{x})}^2\right]d\mathrm{x}=2\mathrm{\pi } \end{gathered}$$

Đáp án C.

$$\begin{gathered} \mathrm{V}\left(\mathrm{t}\right)=3\mathrm{t}-\frac{3{\mathrm{t}}^2}{40}\Rightarrow \mathrm{V}\text{'}\left(\mathrm{t}\right)=3-\frac{6}{40}\mathrm{t} \\ \mathrm{V}\text{'}\left(\mathrm{t}\right)=0\iff \mathrm{t}=20 \\ \Rightarrow {\mathrm{V}}_{\max }=\mathrm{V}\left(20\right)=30 (\mathrm{m}/\mathrm{s}) \\ \Rightarrow \mathrm{S}={\int }_0^{20}\left(3\mathrm{t}-\frac{3{\mathrm{t}}^2}{40}\right)d\mathrm{t}=400\left(\mathrm{m}\right) \end{gathered}$$