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

The integral from 2 to 5 of e^{5x} is (e^{25}-e^{10})/5

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

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

\frac{e ^{25} - e ^{10}}{5}

Decimal Notation

14400975462.2

Solution steps

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

Apply u-substitution

: \int_{10}^{25} e^{u} \frac{1}{5} d u

= \int_{10}^{25} 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_{10}^{25} e^{u} d u

Use the common integral:

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

= \frac{1}{5} [e^{u}]_{10}^{25}

Compute the boundaries:

e^{25} - e^{10}

= \frac{1}{5} (e^{25} - e^{10})

Simplify

= \frac{e ^{25} - e ^{10}}{5}

x \to 1 lim (3 ln (x) + 4)

\int - t + 2 d t

\frac{d y}{d x} = \frac{- 4 sin ( x ) cos ( x )}{y}

\int v (v^{2} + 2)^{2} d v

y^{^{''}} - 6 y^{^{'}} + 25 y = 0

What is the integral from 2 to 5 of e^{5x} ?
The integral from 2 to 5 of e^{5x} is (e^{25}-e^{10})/5