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

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

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

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

\frac{e ^{6} - 1}{3}

Decimal Notation

134.14293 \dots

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Compute the boundaries:

e^{6} - 1

= \frac{1}{3} (e^{6} - 1)

Simplify

= \frac{e ^{6} - 1}{3}

\int (5 x^{4} - 3 x^{2} + 2 x + 6) d x

\int_{0}^{1} \frac{1}{v} d v

\int 1 + sec^{2} (x) d x

sec^{2} (x) - 5

n = 0 \sum \infty 0. 2^{n}

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