What is the integral from 0 to 2 of 15e^{3x} ?

The integral from 0 to 2 of 15e^{3x} is 5(e^6-1)

Popular >

integral from 0 to 2 of 15e^{3x}

\int_{0}^{2} 15 e^{3 x} d x

5 (e^{6} - 1)

Decimal Notation

2012.14396 \dots

Solution steps

\int_{0}^{2} 15 e^{3 x} d x

Take the constant out:

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

= 15 \cdot \int_{0}^{2} e^{3 x} d x

Apply u-substitution

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

= 15 \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

= 15 \cdot \frac{1}{3} \cdot \int_{0}^{6} e^{u} d u

Use the common integral:

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

= 15 \cdot \frac{1}{3} [e^{u}]_{0}^{6}

Simplify

15 \cdot \frac{1}{3} [e^{u}]_{0}^{6} : 5 [e^{u}]_{0}^{6}

= 5 [e^{u}]_{0}^{6}

Compute the boundaries:

e^{6} - 1

= 5 (e^{6} - 1)

\int_{0}^{π} - sin (\frac{n x}{2}) d x

\int_{1}^{2} \frac{1}{y} d y

\int_{2}^{3} (x^{4} - 3 x + 1) d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} ∣ sin (x) ∣ d x

\int_{- 6}^{4} (- \frac{3}{2} x + 1) d x

What is the integral from 0 to 2 of 15e^{3x} ?
The integral from 0 to 2 of 15e^{3x} is 5(e^6-1)