What is the integral of 24e^{20t} ?

The integral of 24e^{20t} is 6/5 e^{20t}+C

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integral of 24e^{20t}

\int 24 e^{20 t} d t

\frac{6}{5} e^{20 t} + C

Solution steps

\int 24 e^{20 t} d t

Take the constant out:

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

= 24 \cdot \int e^{20 t} d t

Apply u-substitution

: \int e^{u} \frac{1}{20} d u

= 24 \cdot \int e^{u} \frac{1}{20} d u

Take the constant out:

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

= 24 \cdot \frac{1}{20} \cdot \int e^{u} d u

Use the common integral:

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

= 24 \cdot \frac{1}{20} e^{u}

Substitute back

u = 20 t

= 24 \cdot \frac{1}{20} e^{20 t}

Simplify

24 \cdot \frac{1}{20} e^{20 t} : \frac{6}{5} e^{20 t}

= \frac{6}{5} e^{20 t}

Add a constant to the solution

= \frac{6}{5} e^{20 t} + C

x \to - 3 - lim (\frac{2}{x ^{2} - 9})

y = x^{5} + cos^{5} (x)

x \to 90 + lim (tan (x))

\frac{d y}{d x} + (\frac{5}{x}) y = - x + 5