What is the integral from 0 to 2 of 6e^{3x} ?

The integral from 0 to 2 of 6e^{3x} is 2(e^6-1)

integral from 0 to 2 of 6e^{3x}

\int_{0}^{2} 6 e^{3 x} d x

2 (e^{6} - 1)

Decimal

804.85758 \dots

Solution steps

\int_{0}^{2} 6 e^{3 x} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{2} e^{3 x} d x

Apply u-substitution

= 6 \cdot \int_{0}^{6} 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

= 6 \cdot \frac{1}{3} \cdot \int_{0}^{6} e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{3} [e^{u}]_{0}^{6}

Simplify

6 \cdot \frac{1}{3} [e^{u}]_{0}^{6} : 2 [e^{u}]_{0}^{6}

= 2 [e^{u}]_{0}^{6}

Compute the boundaries:

e^{6} - 1

= 2 (e^{6} - 1)

\int_{2}^{4} x^{3} - 6 x^{2} + 8 x d x

\int_{- 1}^{3} (3 x^{2} + 5 x - 1) d x

\int_{0}^{\infty} \frac{2}{x ^{2} + 1} d x

\int_{0}^{t} u sin (u) d u

\int_{4}^{7} - 9 lo g_{3} (t) d t

What is the integral from 0 to 2 of 6e^{3x} ?
The integral from 0 to 2 of 6e^{3x} is 2(e^6-1)