What is the integral of 0.06*e^{t/(300)} ?

The integral of 0.06*e^{t/(300)} is 18e^{t/(300)}+C

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integral of 0.06*e^{t/(300)}

\int 0.06 \cdot e^{\frac{t}{300}} d t

18 e^{\frac{t}{300}} + C

Solution steps

\int 0.06 e^{\frac{t}{300}} d t

Take the constant out:

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

= 0.06 \cdot \int e^{\frac{t}{300}} d t

Apply u-substitution

: \int e^{u} \cdot 300 d u

= 0.06 \cdot \int e^{u} \cdot 300 d u

Take the constant out:

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

= 0.06 \cdot 300 \cdot \int e^{u} d u

Use the common integral:

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

= 0.06 \cdot 300 e^{u}

Substitute back

u = \frac{t}{300}

= 0.06 \cdot 300 e^{\frac{t}{300}}

Simplify

= 18 e^{\frac{t}{300}}

Add a constant to the solution

= 18 e^{\frac{t}{300}} + C

\frac{d}{d x} (ln (\frac{2 x + 1}{x - 1}))

n = 1 \sum \infty \frac{7}{( n + 1 ) !}

\frac{d}{d x} (x^{2} 6 x - 7)

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

f (x) = 1 + 5 e^{x}, at x = 1

What is the integral of 0.06*e^{t/(300)} ?
The integral of 0.06*e^{t/(300)} is 18e^{t/(300)}+C