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

The integral from-infinity to 1 of-e^x is -e

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integral from-infinity to 1 of-e^x

\int_{- \infty}^{1} - e^{x} d x

- e

Decimal

- 2.71828 \dots

Solution steps

\int_{- \infty}^{1} - e^{x} d x

Compute the indefinite integral:

\int - e^{x} d x = - e^{x} + C

Compute the boundaries:

\int_{- \infty}^{1} - e^{x} d x = - e - 0

= - e - 0

Simplify

= - e

in v erse l a pl a ce \frac{6 ( s + 1 )}{s ^{2} ( s + 1 )}

\frac{\partial}{\partial x} (arcsin (2 x))

a re a x^{\frac{1}{2}}, x^{2}

d er i v a t i v e y = \frac{9}{x} + 2 sin (x)

d er i v a t i v e p (x) = x - 4 x

What is the integral from-infinity to 1 of-e^x ?
The integral from-infinity to 1 of-e^x is -e