What is the integral of e^{-2t} ?

The integral of e^{-2t} is -1/2 e^{-2t}+C

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integral of e^{-2t}

\int e^{- 2 t} d t

- \frac{1}{2} e^{- 2 t} + C

Solution steps

\int e^{- 2 t} d t

Apply u-substitution

: \int - \frac{1}{2} e^{u} d u

= \int - \frac{1}{2} e^{u} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{2} e^{u}

Substitute back

u = - 2 t

= - \frac{1}{2} e^{- 2 t}

Add a constant to the solution

= - \frac{1}{2} e^{- 2 t} + C

y^{^{'}} - \frac{8}{x} y = \frac{y ^{4}}{x ^{17}}, y (1) = 1

x \to 3 + lim (- \frac{2}{x ^{2} - 9})

\int 5 e^{- 0.4 x} d x

f (x) = 7 sec (x), at x = \frac{π}{6}

x y^{^{'}} + y = x