What is the integral from 0 to 2 of e^{5x} ?

The integral from 0 to 2 of e^{5x} is (e^{10}-1)/5

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integral from 0 to 2 of e^{5x}

\int_{0}^{2} e^{5 x} d x

\frac{e ^{10} - 1}{5}

Decimal Notation

4405.09315 \dots

Solution steps

\int_{0}^{2} e^{5 x} d x

Apply u-substitution

: \int_{0}^{10} e^{u} \frac{1}{5} d u

= \int_{0}^{10} e^{u} \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int_{0}^{10} e^{u} d u

Use the common integral:

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

= \frac{1}{5} [e^{u}]_{0}^{10}

Compute the boundaries:

e^{10} - 1

= \frac{1}{5} (e^{10} - 1)

Simplify

= \frac{e ^{10} - 1}{5}

x \to 0 lim (\frac{x + 1}{x})

\int \frac{10 x ^{2} - 9 x + 4}{x ^{3} + 4 x} d x

\frac{d}{d x} (\frac{7 x ^{2}}{( x ^{2} + x ) ^{3}})

n = 0 \sum \infty \frac{x ^{n}}{3 + 2 x}

\frac{\partial}{\partial x} (ln (x^{2} + y^{2} + 3))

What is the integral from 0 to 2 of e^{5x} ?
The integral from 0 to 2 of e^{5x} is (e^{10}-1)/5