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

Take the constant out:

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

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

Apply u-substitution

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

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

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

\frac{\partial}{\partial y} (cos (x^{3} y^{3}))

\frac{\partial}{\partial x} (- 2 x y sin (x y^{2}))

tan^{2} (x)

y^{^{'}} = y - x^{2} + 1, y (0) = 0