Đáp án:
\(\begin{array}{l}
3)\quad f\kern-2pt\left(\dfrac12\right) =-2\\
4)\quad S = \dfrac{9\ln3}{2} - 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
3)\\
\quad 2xf(x) + x^2f'(x) = 1\\
\Leftrightarrow (x^2)'f(x) + x^2f'(x) = 1\\
\Leftrightarrow \left[x^2f(x)\right]' = 1\\
\Leftrightarrow x^2f(x) = \displaystyle\int 1dx\\
\Leftrightarrow x^2f(x) = x + C\\
\Leftrightarrow f(x) = \dfrac1x + \dfrac{C}{x^2}\\
\text{Ta lại có:}\\
\quad f(1) = 0\\
\Leftrightarrow 1 + C = 0\\
\Leftrightarrow C = -1\\
\text{Ta được:}\\
\quad f(x) = \dfrac1x - \dfrac{1}{x^2}\\
\text{Khi đó:}\\
\quad f\kern-2pt\left(\dfrac12\right) = \dfrac{1}{\dfrac{1}{2}} - \dfrac{1}{\left(\dfrac12\right)^2}\\
\Leftrightarrow f\kern-2pt\left(\dfrac12\right) =-2\\
4)\\
\quad S = \displaystyle\int\limits_1^3 x\ln xdx\\
Đặt\ \begin{cases}u = \ln x\\dv = xdx\end{cases}\Rightarrow \begin{cases}du = \dfrac1xdx\\v = \dfrac{x^2}{2}\end{cases}\\
\text{Ta được:}\\
\quad S = \dfrac{x^2\ln x}{2}\Bigg|_1^3 - \dfrac12\displaystyle\int\limits_1^3xdx\\
\Leftrightarrow S = \dfrac{x^2\ln x}{2}\Bigg|_1^3 - \dfrac{x^2}{4}\Bigg|_1^3\\
\Leftrightarrow S = \dfrac{9\ln3 - 1\ln1}{2} - \dfrac{9 - 1}{4}\\
\Leftrightarrow S = \dfrac{9\ln3}{2} - 2
\end{array}\)
