f=x^-1
∫f2f(f)dx\int_{f}^{2f}(f) dx∫f2f(f)dx
asked by guest on Nov 27, 2024 at 10:42 am
MathBot Answer:
∫f2ffdx=−log(1x)+log(2x)\int_{f}^{2 f}{f}dx = - \log{\left(\frac{1}{x} \right)} + \log{\left(\frac{2}{x} \right)}∫f2ffdx=−log(x1)+log(x2)
∫f2ff dx=∫1x2⋅1x1x dx=∫1x2x1x dx=−log(1x)+log(2x)\begin{aligned}\int\limits_{f}^{2 f} f\, dx&=\int\limits_{\frac{1}{x}}^{2 \cdot \frac{1}{x}} \frac{1}{x}\, dx\\&=\int\limits_{\frac{1}{x}}^{\frac{2}{x}} \frac{1}{x}\, dx\\&=- \log{\left(\frac{1}{x} \right)} + \log{\left(\frac{2}{x} \right)}\end{aligned}f∫2ffdx=x1∫2⋅x1x1dx=x1∫x2x1dx=−log(x1)+log(x2)
