What is the integral of e^{0.5t} ?

The integral of e^{0.5t} is 2e^{0.5t}+C

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integral of e^{0.5t}

\int e^{0.5 t} d t

2 e^{0.5 t} + C

Solution steps

\int e^{0.5 t} d t

Apply u-substitution

: \int e^{u} \frac{1}{0.5} d u

= \int e^{u} \frac{1}{0.5} d u

Take the constant out:

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

= \frac{1}{0.5} \cdot \int e^{u} d u

Simplify

= 2 \cdot \int e^{u} d u

Use the common integral:

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

= 2 e^{u}

Substitute back

u = 0.5 t

= 2 e^{0.5 t}

Add a constant to the solution

= 2 e^{0.5 t} + C

x \to 2 lim (\frac{( 3 x - 6 )}{1 - 4 x - 7})

\frac{\partial}{\partial x} (y^{3} + x^{2} y - 3 y)

\frac{\partial}{\partial y} (x^{2} y^{3})

\int \frac{1}{( x + 2 ) ( x ^{2} + 1 )} d x

f (x) = x - 2, at x = 3