What is the integral from-1 to 1 of |3^x-2^x| ?

The integral from-1 to 1 of |3^x-2^x| is (8ln(2)-3ln(3))/(6ln(3)ln(2))

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integral from-1 to 1 of |3^x-2^x|

\int_{- 1}^{1} ∣ 3^{x} - 2^{x} ∣ d x

\frac{8 ln ( 2 ) - 3 ln ( 3 )}{6 ln ( 3 ) ln ( 2 )}

Decimal Notation

0.49230 \dots

Solution steps

\int_{- 1}^{1} ∣ 3^{x} - 2^{x} ∣ d x

Eliminate Absolutes

= \int_{- 1}^{0} - 3^{x} + 2^{x} d x + \int_{0}^{1} 3^{x} - 2^{x} d x

\int_{- 1}^{0} - 3^{x} + 2^{x} d x = - \frac{2}{3 ln ( 3 )} + \frac{1}{2 ln ( 2 )}

\int_{0}^{1} 3^{x} - 2^{x} d x = \frac{2}{ln ( 3 )} - \frac{1}{ln ( 2 )}

= - \frac{2}{3 ln ( 3 )} + \frac{1}{2 ln ( 2 )} + \frac{2}{ln ( 3 )} - \frac{1}{ln ( 2 )}

Simplify

- \frac{2}{3 ln ( 3 )} + \frac{1}{2 ln ( 2 )} + \frac{2}{ln ( 3 )} - \frac{1}{ln ( 2 )} : \frac{8 ln ( 2 ) - 3 ln ( 3 )}{6 ln ( 3 ) ln ( 2 )}

= \frac{8 ln ( 2 ) - 3 ln ( 3 )}{6 ln ( 3 ) ln ( 2 )}

\int_{\frac{π}{2}}^{0} sin (x) d x

\int_{0}^{\frac{π}{2}} cos (2 u) d u

\int_{1}^{e^{4}} \frac{ln ^{2} ( x ^{4} )}{x} d x

\int_{0}^{1} \frac{x ^{3}}{1 + x} d x

\int_{- \frac{2 π}{3}}^{0} sin (x) d x

What is the integral from-1 to 1 of |3^x-2^x| ?
The integral from-1 to 1 of |3^x-2^x| is (8ln(2)-3ln(3))/(6ln(3)ln(2))