What is the integral from 0 to 5 of 12e^{3x} ?

The integral from 0 to 5 of 12e^{3x} is 4(e^{15}-1)

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integral from 0 to 5 of 12e^{3x}

\int_{0}^{5} 12 e^{3 x} d x

4 (e^{15} - 1)

Decimal

13076065.48988 \dots

Solution steps

\int_{0}^{5} 12 e^{3 x} d x

Take the constant out:

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

= 12 \cdot \int_{0}^{5} e^{3 x} d x

Apply u-substitution

: \int_{0}^{15} e^{u} \frac{1}{3} d u

= 12 \cdot \int_{0}^{15} e^{u} \frac{1}{3} d u

Take the constant out:

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

= 12 \cdot \frac{1}{3} \cdot \int_{0}^{15} e^{u} d u

Use the common integral:

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

= 12 \cdot \frac{1}{3} [e^{u}]_{0}^{15}

Simplify

12 \cdot \frac{1}{3} [e^{u}]_{0}^{15} : 4 [e^{u}]_{0}^{15}

= 4 [e^{u}]_{0}^{15}

Compute the boundaries:

e^{15} - 1

= 4 (e^{15} - 1)

\frac{\partial}{\partial z} (x sin (y) sin (z))

x y - 2 y^{2} = 4, (9, 4)

ln (\frac{x}{x ^{2} + 1})

\frac{3 s + 1}{s ^{2} ( s ^{2} + 4 )}

x y^{^{'}} - y = \frac{x ^{2}}{y ^{2}}

What is the integral from 0 to 5 of 12e^{3x} ?
The integral from 0 to 5 of 12e^{3x} is 4(e^{15}-1)