What is the integral of e^{-3/2 t} ?

The integral of e^{-3/2 t} is -2/3 e^{-(3t)/2}+C

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integral of e^{-3/2 t}

\int e^{- \frac{3}{2} t} d t

- \frac{2}{3} e^{- \frac{3 t}{2}} + C

Solution steps

\int e^{- \frac{3}{2} t} d t

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \int e^{- \frac{3 t}{2}} d t

Apply u-substitution

: \int - \frac{2}{3} e^{u} d u

= \int - \frac{2}{3} e^{u} d u

Take the constant out:

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

= - \frac{2}{3} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{2}{3} e^{u}

Substitute back

u = - \frac{3 t}{2}

= - \frac{2}{3} e^{- \frac{3 t}{2}}

Add a constant to the solution

= - \frac{2}{3} e^{- \frac{3 t}{2}} + C

3 x^{2} + 4 x + 8

(x^{2} + 7)^{5}

x, x^{3} - x^{2} - x

2 sin (x), 4 cos (x), [0, 0.6 π]

\frac{d}{d x} ((x^{5} - x^{3} + 3)^{4})

What is the integral of e^{-3/2 t} ?
The integral of e^{-3/2 t} is -2/3 e^{-(3t)/2}+C