What is the integral of (e^{-t}) ?

The integral of (e^{-t}) is -e^{-t}+C

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integral of (e^{-t})

\int (e^{- t}) d t

- e^{- t} + C

Solution steps

\int e^{- t} d t

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - e^{u}

Substitute back

u = - t

= - e^{- t}

Add a constant to the solution

= - e^{- t} + C

\int_{\frac{1}{2}}^{1} \frac{y + 4}{y ^{2} + y} d y

y = 0.7 (1.02)^{x}

f (x) = 1 - 4 x^{2}

\frac{d y}{d x} = y (0.1 - 1 0^{- 7} y)

\frac{d}{d x} (ln ∣ x - 1 ∣ - ln ∣ x - 3 ∣)