What is the integral from 0 to 1 of 6e^{3x} ?

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

integral from 0 to 1 of 6e^{3x}

\int_{0}^{1} 6 e^{3 x} d x

2 (e^{3} - 1)

Decimal

38.17107 \dots

Solution steps

\int_{0}^{1} 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}^{1} e^{3 x} d x

Apply u-substitution

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

Use the common integral:

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

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

Simplify

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

= 2 [e^{u}]_{0}^{3}

Compute the boundaries:

e^{3} - 1

= 2 (e^{3} - 1)

\int_{0}^{4} 6 - 2 x d x

\int_{- 2}^{5} (3 + x - 2 ∣ x ∣) d x

\int_{1}^{2} 8 x d x

\int_{0}^{3} (6 + 6 y - y^{2}) d y

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

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