What is the integral from-1 to 1 of 7^x ?

The integral from-1 to 1 of 7^x is (48)/(7ln(7))

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integral from-1 to 1 of 7^x

\int_{- 1}^{1} 7^{x} d x

\frac{48}{7 ln ( 7 )}

Decimal Notation

3.52387 \dots

Solution steps

\int_{- 1}^{1} 7^{x} d x

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

= [\frac{7 ^{x}}{ln ( 7 )}]_{- 1}^{1}

Compute the boundaries:

\frac{48}{7 ln ( 7 )}

= \frac{48}{7 ln ( 7 )}

x \to - 1 lim (\frac{x}{x + 1})

\frac{d}{d x} ((ln (3))^{x})

x^{4} - 5 x^{2} + 4, at x = 1

\frac{d y}{d t} = 2 y - y^{2}

\int 1 - sin (2 x) d x

What is the integral from-1 to 1 of 7^x ?
The integral from-1 to 1 of 7^x is (48)/(7ln(7))