What is the integral from 1 to 3 of 8^x-2^x ?

The integral from 1 to 3 of 8^x-2^x is (162)/(ln(2))

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integral from 1 to 3 of 8^x-2^x

\int_{1}^{3} 8^{x} - 2^{x} d x

\frac{162}{ln ( 2 )}

Decimal Notation

233.71659 \dots

Solution steps

\int_{1}^{3} 8^{x} - 2^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} 8^{x} d x - \int_{1}^{3} 2^{x} d x

\int_{1}^{3} 8^{x} d x = \frac{168}{ln ( 2 )}

\int_{1}^{3} 2^{x} d x = \frac{6}{ln ( 2 )}

= \frac{168}{ln ( 2 )} - \frac{6}{ln ( 2 )}

Simplify

\frac{168}{ln ( 2 )} - \frac{6}{ln ( 2 )} : \frac{162}{ln ( 2 )}

= \frac{162}{ln ( 2 )}

(3, 1), (- 2, - 1)

- 7 y^{^{'}} + 21 x^{2} = 0

\frac{d}{d x} (25 e^{- 5 x})

\int (13 - 6 x + 21 x^{6}) d x

\frac{d}{d x} (\frac{cos ( 2 x )}{sin ( x )})

What is the integral from 1 to 3 of 8^x-2^x ?
The integral from 1 to 3 of 8^x-2^x is (162)/(ln(2))