What is the integral of 9e^{-t} ?

The integral of 9e^{-t} is -9e^{-t}+C

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integral of 9e^{-t}

\int 9 e^{- t} d t

- 9 e^{- t} + C

Solution steps

\int 9 e^{- t} d t

Take the constant out:

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

= 9 \cdot \int e^{- t} d t

Apply u-substitution

: \int - e^{u} d u

= 9 \cdot \int - e^{u} d u

Take the constant out:

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

= 9 (- \int e^{u} d u)

Use the common integral:

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

= 9 (- e^{u})

Substitute back

u = - t

= 9 (- e^{- t})

Simplify

= - 9 e^{- t}

Add a constant to the solution

= - 9 e^{- t} + C

x \to \infty lim (- 1 \cdot x)

f (x) = sec^{2} (x) - tan^{2} (x)

\int e^{2 x} \cdot x d x

y^{^{'}} = 0.05 ln (\frac{4000}{y}) y

ln (1 + 2 x)