Đáp án:
\(\begin{array}{l}
1)\quad 1009\\
2)\quad \dfrac{12\,113}{6}\\
3)\quad \ln\dfrac{e + 2018}{2019}\\
4)\quad 2(\sqrt2 - 1)
\end{array}\)
Giải thích các bước giải:
$\begin{array}{l}1)\quad I = \displaystyle\int\limits_0^{\tfrac{\pi}{4}}2018\cos2xdx\\
\to I = 1009\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\cos2xd(2x)\\
\to I = 1009\sin2x\Bigg|_0^{\tfrac{\pi}{4}}\\
\to I = 1009\left(\sin\dfrac{\pi}{2} - \sin0\right)\\
\to I = 1009\\
2)\quad I = \displaystyle\int\limits_0^1(x^2 + x + 2018)dx\\
\to I = \displaystyle\int\limits_0^1x^2dx + \displaystyle\int\limits_0^1xdx + 2018\displaystyle\int\limits_0^1dx\\
\to I = \dfrac{x^3}{3}\Bigg|_0^1 + \dfrac{x^2}{2}\Bigg|_0^1 + 2018x\Bigg|_0^1\\
\to I =\dfrac13 + \dfrac12 + 2018\\
\to I = \dfrac{12\,113}{6}\\
3)\quad I = \displaystyle\int\limits_0^1\dfrac{e^xdx}{2018 + e^x}\\
\to I = \displaystyle\int\limits_0^1\dfrac{d(e^x + 2018)}{e^x + 2018}\\
\to I = \ln|e^x + 2018|\Bigg|_0^1\\
\to I = \ln(e + 2018) - \ln2019\\
\to I = \ln\dfrac{e + 2018}{2019}\\
4)\quad I = \displaystyle\int\limits_1^e\dfrac{dx}{x\sqrt{1 + \ln x}}\\
\to I = \displaystyle\int\limits_1^e\dfrac{d(1 + \ln x)}{\sqrt{1 + \ln x}}\\
\to I = 2\sqrt{1 + \ln x}\Bigg|_1^e\\
\to I = 2\left(\sqrt{1 + \ln e} - \sqrt{1 + \ln1}\right)\\
\to I = 2(\sqrt2 - 1)
\end{array}$
