f=x^-1

f2f(f)dx\int_{f}^{2f}(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=1x21x1xdx=1x2x1xdx=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}