Xét tích phân $I=\displaystyle\displaystyle\int\limits_1^{\rm{e}^2}\dfrac{\left(1+2\ln x\right)^2}{x}\mathrm{\,d}x$, nếu đặt $t=1+2\ln{x}$ thì $I$ bằng
 | $\dfrac{1}{2}\displaystyle\displaystyle\int\limits_1^{e^2}t^2\mathrm{\,d}t$ |
 | $2\displaystyle\displaystyle\int\limits_1^5t^2\mathrm{\,d}t$ |
 | $2\displaystyle\displaystyle\int\limits_1^{e^2}t^2\mathrm{\,d}t$ |
 | $\dfrac{1}{2}\displaystyle\displaystyle\int\limits_1^5t^2\mathrm{\,d}t$ |
Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ thỏa $\displaystyle\displaystyle\int\limits_{0}^{1}f(x)\mathrm{d}x=2$ và $\displaystyle\displaystyle\int\limits_{0}^2f(3x+1)\mathrm{d}x=6$. Tính $I=\displaystyle\displaystyle\int\limits_{0}^{7}f(x)\mathrm{d}x$.
| $I=20$ |
| $I=8$ |
| $I=18$ |
| $I=16$ |
Cho $\displaystyle\displaystyle\int\limits_{4}^{9}f(x)\mathrm{d}x=10$. Tính tích phân $J=\displaystyle\displaystyle\int\limits_{0}^{1}f(5x+4)\mathrm{d}x$.
| $J=2$ |
| $J=10$ |
| $J=50$ |
| $J=4$ |
Biết $f\left(x\right)$ là hàm số liên tục trên $\mathbb{R}$ và $\displaystyle\displaystyle\int\limits_{0}^{9}f\left(x\right)\mathrm{d}x=9$. Khi đó tính $I=\displaystyle\displaystyle\int\limits_{2}^{5}f\left(3x-6\right)\mathrm{d}x$.
| $I=27$ |
| $I=24$ |
| $I=3$ |
| $I=0$ |
Nếu $\displaystyle\displaystyle\int\limits_1^3f(x)\mathrm{\,d}x=3$ thì $\displaystyle\displaystyle\int\limits_1^5f\left(\dfrac{x+1}{2}\right)\mathrm{\,d}x$ bằng
| $\dfrac{3}{2}$ |
| $3$ |
| $\dfrac{5}{2}$ |
| $6$ |
Bằng cách đổi biến số $t=1+\ln x$ thì tích phân $\displaystyle\displaystyle\int\limits_1^\mathrm{e}\dfrac{(1+\ln x)^2}{x}\mathrm{\,d}x$ trở thành
| $\displaystyle\displaystyle\int\limits_1^\mathrm{e}t^2\mathrm{\,d}t$ |
| $\displaystyle\displaystyle\int\limits_1^2t^2\mathrm{\,d}t$ |
| $\displaystyle\displaystyle\int\limits_1^4t^2\mathrm{\,d}t$ |
| $\displaystyle\displaystyle\int\limits_1^2(1+t)^2\mathrm{\,d}t$ |
Biết $\displaystyle\displaystyle\int\limits_0^1x\sqrt{x^2+4}\mathrm{\,d}x=\dfrac{1}{a}\left(\sqrt{b^3}-c\right)$. Tính $Q=abc$.
| $Q=120$ |
| $Q=15$ |
| $Q=-120$ |
| $Q=40$ |
Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ và $\displaystyle\displaystyle\int\limits_{0}^{4}f(x)\mathrm{\,d}x=2020$. Giá trị của $\displaystyle\displaystyle\int\limits_{0}^{2}xf\left(x^2\right)\mathrm{\,d}x$ bằng
| $1008$ |
| $4040$ |
| $1010$ |
| $2019$ |
Nếu đặt $u=2x+1$ thì $\displaystyle\displaystyle\int\limits_{0}^{1}(2x+1)^4\mathrm{\,d}x$ bằng
| $\dfrac{1}{2}\displaystyle\displaystyle\int\limits_{1}^{3}u^4\mathrm{\,d}u$ |
| $\displaystyle\displaystyle\int\limits_{1}^{3}u^4\mathrm{\,d}u$ |
| $\dfrac{1}{2}\displaystyle\displaystyle\int\limits_{0}^{1}u^4\mathrm{\,d}u$ |
| $\displaystyle\displaystyle\int\limits_{0}^{1}u^4\mathrm{\,d}u$ |
Tính $I=\displaystyle\displaystyle\int\limits_{0}^{a}\dfrac{x^3+x}{\sqrt{x^2+1}}\mathrm{\,d}x$.
| $I=\left(a^2+1\right)\sqrt{a^2+1}+1$ |
| $I=\left(a^2+1\right)\sqrt{a^2+1}-1$ |
| $I=\dfrac{1}{3}\left[\left(a^2+1\right)\sqrt{a^2+1}-1\right]$ |
| $I=\dfrac{1}{3}\left[\left(a^2+1\right)\sqrt{a^2+1}+1\right]$ |
Cho hàm số $f(x)=\begin{cases} x^2-1 &\text{khi }x\geq2\\ x^2-2x+3 &\text{khi }x< 2 \end{cases}$. Tích phân $\displaystyle\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}f\left(2\sin x+1\right)\cos x\mathrm{\,d}x$ bằng
| $\dfrac{23}{3}$ |
| $\dfrac{23}{6}$ |
| $\dfrac{17}{6}$ |
| $\dfrac{17}{3}$ |
Cho hàm số \(y=f(x)\) có đạo hàm liên tục trên \(\mathbb{R}\) và thỏa mãn \(f(2)=16\), \(\displaystyle\int\limits_{0}^{2}f(x)\mathrm{\,d}x=4\). Tính \(I=\displaystyle\int\limits_{0}^{1}xf'(2x)\mathrm{\,d}x\).
| \(I=13\) |
| \(I=20\) |
| \(I=12\) |
| \(I=7\) |
Tính tích phân \(I=\displaystyle\int\limits_{1}^{\mathrm{e}}\dfrac{\sqrt{2+\ln x}}{2x}\mathrm{\,d}x\).
| \(\dfrac{3\sqrt{3}+2\sqrt{2}}{3}\) |
| \(\dfrac{\sqrt{3}+\sqrt{2}}{3}\) |
| \(\dfrac{\sqrt{3}-\sqrt{2}}{3}\) |
| \(\dfrac{3\sqrt{3}-2\sqrt{2}}{3}\) |
Nếu \(t=\sqrt{x^2+3}\) thì tích phân \(I=\displaystyle\int\limits_{1}^{2}x\sqrt{x^2+3}\mathrm{\,d}x\) trở thành
| \(I=\displaystyle\int\limits_{2}^{\sqrt{7}}t\mathrm{\,d}t\) |
| \(I=\displaystyle\int\limits_{2}^{7}t^2\mathrm{\,d}t\) |
| \(I=\displaystyle\int\limits_{2}^{\sqrt{7}}t^2\mathrm{\,d}t\) |
| \(I=\displaystyle\int\limits_{2}^{\sqrt{7}}t^3\mathrm{\,d}t\) |
Xét \(\displaystyle\int\limits_0^2x\cdot\mathrm{e}^{x^2}\mathrm{\,d}x\), nếu đặt \(u=x^2\) thì \(\displaystyle\int\limits_0^2x\cdot\mathrm{e}^{x^2}\mathrm{\,d}x\) bằng
| \(2\displaystyle\int\limits_0^2\mathrm{e}^u\mathrm{\,d}u\) |
| \(2\displaystyle\int\limits_0^4\mathrm{e}^u\mathrm{\,d}u\) |
| \(\dfrac{1}{2}\displaystyle\int\limits_0^2\mathrm{e}^u\mathrm{\,d}u\) |
| \(\dfrac{1}{2}\displaystyle\int\limits_0^4\mathrm{e}^u\mathrm{\,d}u\) |
Hàm số \(y=f(x)\) liên tục trên \([1;4]\) và thỏa mãn \(f(x)=\dfrac{f\left(2\sqrt{x}-1\right)}{\sqrt{x}}+\dfrac{\ln x}{x}\). Tính tích phân \(I=\displaystyle\int\limits_{3}^{4}f(x)\mathrm{\,d}x\).
| \(I=3+2\ln^22\) |
| \(I=\ln^2\) |
| \(I=2\ln2\) |
| \(I=2\ln^22\) |
Cho \(\displaystyle\int\limits_{-1}^5f(x)\mathrm{\,d}x=9\). Tính \(I=\displaystyle\int\limits_0^2f(3x-1)\mathrm{\,d}x\).
| \(I=26\) |
| \(I=9\) |
| \(I=3\) |
| \(I=27\) |
Cho \(I=\displaystyle\int\limits_0^4x\sqrt{1+2x}\mathrm{\,d}x\) và \(u=\sqrt{2x+1}\). Mệnh đề nào sau đây sai?
| \(I=\dfrac{1}{2}\left(\dfrac{u^5}{5}-\dfrac{u^3}{3}\right)\bigg|_1^3\) |
| \(I=\displaystyle\int\limits_1^3u^2\left(u^2-1\right)\mathrm{\,d}u\) |
| \(I=\dfrac{1}{2}\displaystyle\int\limits_1^3x^2\left(x^2-1\right)\mathrm{\,d}x\) |
| \(I=\dfrac{1}{2}\displaystyle\int\limits_1^3u^2\left(u^2-1\right)\mathrm{\,d}u\) |
Cho tích phân \(I=\displaystyle\int_0^4x\sqrt{x^2+9}\mathrm{\,d}x\). Khi đặt \(t=\sqrt{x^2+9}\) thì tích phân đã cho trở thành
| \(I=\displaystyle\int_3^5t\mathrm{\,d}t\) |
| \(I=\displaystyle\int_0^4t\mathrm{\,d}t\) |
| \(I=\displaystyle\int_0^4t^2\mathrm{\,d}t\) |
| \(I=\displaystyle\int_3^5t^2\mathrm{\,d}t\) |
Cho \(\displaystyle\int\limits_1^3 \dfrac{\left(x+6\right)^{2017}}{x^{2019}}\mathrm{\,d}x=\dfrac{a^{2018}-3^{2018}}{6\cdot 2018}\). Tính \(a\).
| \(7\) |
| \(9\) |
| \(6\) |
| \(8\) |