Lời giải

* \( 2^x - 2^{-x} > 0 \iff 2^x > 2^{-x} \iff x >-x \iff x > 0 \)

\( \int_{-1}^1 |2^x - 2^{-x}| \, dx = \int_{-1}^0 |2^x - 2^{-x}| \, dx + \int_0^1 |2^x - 2^{-x}| \, dx \)

\(  = \int_{-1}^0 (2^{-x} - 2^x) \, dx + \int_0^1 (2^x - 2^{-x}) \, dx \)

\( = \left( -\frac{2^{-x}}{\ln 2} - \frac{2^x}{\ln 2} \right) \bigg|_{-1}^0 + \left( \frac{2^x}{\ln 2} + \frac{2^{-x}}{\ln 2} \right) \bigg|_0^1 \)

\(  = \left( -\frac{2}{\ln 2} + \frac{2}{\ln 2} + \frac{1}{2\ln 2} \right) + \left( \frac{2}{\ln2} + \frac{1}{2\ln 2} \ -\frac{2}{\ln 2} \right) \)

\(  = \frac{1}{\ln 2}\)  (Bấm máy= 1.442695)

* Bấm:  \( \int_{-1}^1 \sqrt{(2^x - 2^{-x})^2} \, dx = 1.442695 \)