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

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

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

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

- \frac{1}{e ^{t}} + 1

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{e ^{t}} + 1

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

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