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

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

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integral of-2e^{2t}

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

- e^{2 t} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - 2 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = 2 t

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

Simplify

- 2 \cdot \frac{1}{2} e^{2 t} : - e^{2 t}

= - e^{2 t}

Add a constant to the solution

= - e^{2 t} + C

\int \frac{1}{1 + 17 x} d x

\int (x sin (x)) d x

y = \frac{2 x}{4 + x ^{2}}

\frac{d x}{d t} = \frac{2}{x}

x \to 6 lim ((x + 1)^{2})