Tính tích phân sau: \(I=\int\limits^{\frac{\pi}{2}}_0\left(x+1\right)\sin x\text{d}x\) \(I=1\) \(I=2\) \(I=\pi\) \(I=\frac{\pi}{2}\) Hướng dẫn giải: Dùng phương pháp tích phân từng phần (\(\int uv'=uv-\int u'v\)) \(\begin{cases}u=x+1\\v'=\sin x\end{cases}\) => \(\begin{cases}u'=1\\v=-\cos x\end{cases}\) Suy ra: \(I=\int\limits^{\frac{\pi}{2}}_0\left(x+1\right)\sin x\text{d}x=-\left(x+1\right)\cos x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\left(-\cos x\right)\text{d}x\) \(=-\left(x+1\right)\cos x|^{\frac{\pi}{2}}_0+\sin x|^{\frac{\pi}{2}}_0\) \(=2\) Đáp số : \(I=2\).
Tính tích phân \(I=\int\limits^e_1x^2\ln x\text{d}x\) \(I=\frac{1}{3}\) \(I=\frac{2}{9}e^3\) \(I=\frac{1}{3}e^3\) \(I=\frac{2}{9}e^3+\frac{1}{9}\) Hướng dẫn giải: Sử dụng công thức tính tích phân từng phần \(\int uv'=uv-\int u'v\) \(\begin{cases}u=\ln x\\v'=x^2\end{cases}\) => \(\begin{cases}u'=\frac{1}{x}\\v=\frac{1}{3}x^3\end{cases}\) \(I=\int\limits^e_1x^2\ln x\text{d}x=\frac{1}{3}x^3\ln x|^e_1-\frac{1}{3}\int\limits^e_1x^2\text{d}x\) \(=\frac{1}{3}x^3\ln x|^e_1-\frac{1}{3}.\frac{1}{3}x^3|^e_1\) \(=\frac{2}{9}e^3+\frac{1}{9}\) Đáp số:\(I=\frac{2}{9}e^3+\frac{1}{9}\) .
Tính \(I=\int\limits^1_0\ln\left(1+x\right)\text{d}x\) \(I=2\ln2-1\) \(I=2\ln2+1\) \(I=1\) \(I=\ln2\) Hướng dẫn giải: Sử dụng công thức tính tích phân từng phần \(\int uv'=uv-\int u'v\) \(\begin{cases}u=\ln\left(1+x\right)\\v'=1\end{cases}\) => \(\begin{cases}u'=\frac{1}{1+x}\\v=x\end{cases}\) \(I=\int\limits^1_0\ln\left(1+x\right)\text{d}x=x.\ln\left(1+x\right)|^1_0-\int\limits^1_0\frac{x}{1+x}\text{d}x\) \(=x.\ln\left(1+x\right)|^1_0-\int\limits^1_0\left(1-\frac{1}{x+1}\right)\text{d}x\) \(=x.\ln\left(1+x\right)|^1_0-\left[x-\ln\left(1+x\right)\right]|^1_0\) \(=2\ln2-1\) Đáp số: \(I=2\ln2-1\).
Tính tích phân \(I=\int\limits^1_0\left(x^2-2x-1\right)e^{-x}\text{d}x\) \(I=1\) \(I=-1\) \(I=e\) \(I=\frac{1}{e}\) Hướng dẫn giải: Cách 1: Ta tính các nguyên hàm: \(I_1=\int x^2e^{-x}\text{d}x\) , \(I_2=\int xe^{-x}\text{d}x\), \(I_3=\int e^{-x}\text{d}x\) Ta có: \(I_3=\int e^{-x}\text{d}x=-\int e^{-x}\text{d}\left(-x\right)=-e^{-x}\) Để tính \(I_2\), ta đặt: \(\begin{cases}u=x\\v'=e^{-x}\end{cases}\) => \(\begin{cases}u'=1\\v=-e^{-x}\end{cases}\) => \(I_2=\int xe^{-x}\text{d}x=-xe^{-x}+\int e^{-x}\text{d}x=-xe^{-x}-e^{-x}\) Để tính \(I_1\), ta đặt: \(\begin{cases}u=x^2\\v=e^{-x}\end{cases}\) => \(\begin{cases}u'=2x\\v=-e^{-x}\end{cases}\) => \(I_1=-x^2e^{-x}+2\int xe^{-x}\text{d}x=-x^2e^{-x}+2I_2\) \(=-x^2e^{-x}+2\left(-xe^{-x}-e^{-x}\right)\) \(=-x^2e^{-x}-2xe^{-x}-2e^{-x}\) Vậy ta có: \(\int\left(x^2-2x-1\right)e^{-x}\text{d}x=\int x^2e^{-x}\text{d}x-2\int xe^{-x}\text{d}x-\int e^{-x}\text{d}x\) \(=I_1-2I_2-I_3\) \(=\left(-x^2e^{-x}-2xe^{-x}-2e^{-x}\right)-2\left(-xe^{-x}-e^{-x}\right)+e^{-x}\) \(=\left(-x^2+1\right)e^{-x}\) Suy ra: \(I=\left(-x^2+1\right)e^{-x}|^1_0=-1\) . Đáp số: \(I=-1\).
Tính \(I=\int\limits^1_0\left(1+3x\right)^{\frac{3}{2}}\text{d}x\) \(I=\frac{62}{15}\) \(I=4\) \(I=\frac{62}{5}\) \(I=\frac{155}{2}\) Hướng dẫn giải: \(I=\int\limits^1_0\left(1+3x\right)^{\frac{3}{2}}\text{d}x\) \(=\frac{1}{3}\int\limits^1_0\left(1+3x\right)^{\frac{3}{2}}\text{d}\left(1+3x\right)\) \(=\frac{1}{3}.\frac{1}{\frac{3}{2}+1}\left(1+3x\right)^{\frac{3}{2}+1}|^1_0\) \(=\frac{2}{15}\left(1+3x\right)^{\frac{5}{2}}|^1_0\) \(=\frac{62}{15}\) Chọn A.
Tính \(I=\int\limits^{\frac{1}{2}}_0\frac{x^3-1}{x^2-1}\text{d}x\) \(I=\frac{1}{4}+\ln\frac{3}{2}\) \(I=\ln\frac{3}{2}\) \(I=\frac{1}{8}\) \(I=\frac{1}{8}+\ln\frac{3}{2}\) Hướng dẫn giải: \(I=\int\limits^{\frac{1}{2}}_0\frac{x^3-1}{x^2-1}\text{d}x\) \(=\int\limits^{\frac{1}{2}}_0\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)}\text{d}x\) \(=\int\limits^{\frac{1}{2}}_0\frac{x^2+x+1}{x+1}\text{d}x\) \(=\int\limits^{\frac{1}{2}}_0\frac{x\left(x+1\right)+1}{x+1}\text{d}x\) \(=\int\limits^{\frac{1}{2}}_0\left(x+\frac{1}{x+1}\right)\text{d}x\) \(=\int\limits^{\frac{1}{2}}_0x\text{d}x+\int\limits^{\frac{1}{2}}_0\frac{1}{x+1}\text{d}\left(x+1\right)\) \(=\left[\frac{1}{2}x^2+\ln\left(x+1\right)\right]|^{\frac{1}{2}}_0\) \(=\frac{1}{8}+\ln\frac{3}{2}\) Chọn D.
Tính \(I=\int\limits^2_1\frac{\ln\left(1+x\right)}{x^2}\text{d}x\) \(I=3\ln\frac{2\sqrt{3}}{3}\) \(I=2\ln\frac{\sqrt{3}}{3}\) \(I=3\ln2\) \(I=3\ln\sqrt{3}\) Hướng dẫn giải: Đặt: \(\begin{cases}u=\ln\left(1+x\right)\\v'=\frac{1}{x^2}\end{cases}\) => \(\begin{cases}u'=\frac{1}{1+x}\\v=\int x^{-2}\text{d}x=-x^{-1}=-\frac{1}{x}\end{cases}\) \(I=-\frac{\ln\left(1+x\right)}{x}|^2_1+\int\limits^2_1\frac{1}{x\left(x+1\right)}\text{d}x\) Ta có: \(\int\frac{1}{x\left(x+1\right)}\text{d}x=\int\left[\frac{1}{x}-\frac{1}{x+1}\right]\text{d}x=\ln x-\ln\left(x+1\right)\) Suy ra: \(I=\left[-\frac{\ln\left(1+x\right)}{x}+\ln x-\ln\left(1+x\right)\right]|^2_1\) \(=-\frac{3}{2}\ln3+3\ln2=3\ln\frac{2}{\sqrt{3}}=3\ln\frac{2\sqrt{3}}{3}\) Đáp số : \(I=3\ln\frac{2\sqrt{3}}{3}\). Kiểm tra
Cho a < b < c, \(\int\limits^b_af\left(x\right)\text{d}x=5\), \(\int\limits^b_cf\left(x\right)\text{d}x=2\) . Tính \(\int\limits^c_af\left(x\right)\text{d}x\) . \(\int\limits^c_af\left(x\right)\text{d}x=-2\) \(\int\limits^c_af\left(x\right)\text{d}x=3\) \(\int\limits^c_af\left(x\right)\text{d}x=8\) \(\int\limits^c_af\left(x\right)\text{d}x=0\) Hướng dẫn giải: \(\int\limits^c_af\left(x\right)\text{d}x=\int\limits^b_af\left(x\right)\text{d}x+\int\limits^c_bf\left(x\right)\text{d}x\) \(=\int\limits^b_af\left(x\right)\text{d}x-\int\limits^b_cf\left(x\right)\text{d}x=5-2=3\) Chọn B.
Biết hàm số f(x) có đạo hàm \(f'\left(x\right)\) và \(f\left(0\right)=\pi\) , \(\int\limits^{\pi}_0f'\left(x\right)\text{d}x=3\pi\). Tính \(f\left(\pi\right)\). \(f\left(\pi\right)=0\) \(f\left(\pi\right)=-\pi\) \(f\left(\pi\right)=4\pi\) \(f\left(\pi\right)=2\pi\) Hướng dẫn giải: Ta có: \(\int\limits^{\pi}_0f'\left(x\right)\text{d}x=f\left(\pi\right)-f\left(0\right)\), suy ra: \(f\left(\pi\right)-\pi=3\pi\) => \(f\left(\pi\right)=4\pi\). Chọn C.
Trong các khẳng định sau, khẳng định nào sai ? \(\left(\tan x-x\right)'=\tan^2x\) \(\int\limits^{\frac{\pi}{4}}_0x\tan^2xdx=x\left(\tan x-x\right)|^{\frac{\pi}{4}}_0-\int\limits^{\frac{\pi}{4}}_0\left(\tan x-x\right)dx\) \(\int\limits^{\frac{\pi}{4}}_0x\tan^2xdx=\frac{\pi}{4}\left(1-\frac{\pi}{4}\right)+\int\limits^{\frac{\pi}{4}}_0\frac{d\cos x}{\cos x}+\int\limits^{\frac{\pi}{4}}_0xdx\) \(\int\limits^{\frac{\pi}{4}}_0x\tan^2dx=\frac{\pi}{4}+\frac{\pi^2}{32}-\frac{1}{2}\ln2\) Hướng dẫn giải: Ta kiểm tra: *) \(\left(\tan x-x\right)'=\frac{1}{\cos^2x}-1=\frac{1-\cos^2x}{\cos^2x}=\frac{\sin^2x}{\cos^2x}=\tan^2x\) *) Đặt \(\begin{cases}u=x\\v'=\tan^2x\end{cases}\) => \(\begin{cases}u'=1\\v=\int\tan^2x\text{d}x=\tan x-x\end{cases}\) (chú ý dựa vào ý trên \(\left(\tan x-x\right)'=\tan^2x\)) Vậy \(\int\limits^{\frac{\pi}{4}}_0x\tan^2xdx=x\left(\tan x-x\right)|^{\frac{\pi}{4}}_0-\int\limits^{\frac{\pi}{4}}_0\left(\tan x-x\right)dx\) \(=\frac{\pi}{4}\left(1-\frac{\pi}{4}\right)-\int\limits^{\frac{\pi}{4}}_0\frac{-\text{d}\left(\cos x\right)}{\cos x}+\int\limits^{\frac{\pi}{4}}_0x\text{d}x\) \(=\frac{\pi}{4}\left(1-\frac{\pi}{4}\right)+\ln\cos x|^{\frac{\pi}{4}}_0+\frac{x^2}{2}|^{\frac{\pi}{4}}_0\) \(=\frac{\pi}{4}-\frac{\pi^2}{32}-\frac{1}{2}\ln2\) Vậy \(\int\limits^{\frac{\pi}{4}}_0x\tan^2dx=\frac{\pi}{4}+\frac{\pi^2}{32}-\frac{1}{2}\ln2\) là khẳng định sai.
