What is the integral from 0 to 1 of 1/(e^x) ?

The integral from 0 to 1 of 1/(e^x) is -1/e+1

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integral from 0 to 1 of 1/(e^x)

\int_{0}^{1} \frac{1}{e ^{x}} d x

- \frac{1}{e} + 1

Decimal Notation

0.63212 \dots

Solution steps

\int_{0}^{1} \frac{1}{e ^{x}} d x

Apply u-substitution

: \int_{1}^{e} \frac{1}{u ^{2}} d u

= \int_{1}^{e} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{e}

= [- \frac{1}{u}]_{1}^{e}

Compute the boundaries:

- \frac{1}{e} + 1

= - \frac{1}{e} + 1

e^{x^{2} y}

\frac{\partial}{\partial x} (2 ln (x) + y)

x \to 1 lim (e^{9 x^{3} - 9 x})

\frac{d ^{2} y}{d t ^{2}} = 150 y

2 x

What is the integral from 0 to 1 of 1/(e^x) ?
The integral from 0 to 1 of 1/(e^x) is -1/e+1