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

The integral from 2 to-1 of 3^x is -(26)/(3ln(3))

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integral from 2 to-1 of 3^x

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

- \frac{26}{3 ln ( 3 )}

Decimal

- 7.88873 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - [\frac{3 ^{x}}{ln ( 3 )}]_{- 1}^{2}

Compute the boundaries:

\frac{26}{3 ln ( 3 )}

= - \frac{26}{3 ln ( 3 )}

n = 2 \sum \infty \frac{1}{n ^{\frac{3}{2}}}

\int \frac{( x ^{2} )}{( ( x - 1 ) ( x ^{2} + 2 x + 1 ) )} d x

\frac{d}{d x} (\frac{1}{cos ( x )})

\frac{s + 1}{( s + 2 ) ( s + 3 )}

x \to 0 + lim (ln (\frac{7}{x ^{2}}))

What is the integral from 2 to-1 of 3^x ?
The integral from 2 to-1 of 3^x is -(26)/(3ln(3))