What is the integral from 0 to 2 of e^{4x} ?

The integral from 0 to 2 of e^{4x} is (e^8-1)/4

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integral from 0 to 2 of e^{4x}

\int_{0}^{2} e^{4 x} d x

\frac{e ^{8} - 1}{4}

Decimal Notation

744.98949 \dots

Solution steps

\int_{0}^{2} e^{4 x} d x

Apply u-substitution

: \int_{0}^{8} e^{u} \frac{1}{4} d u

= \int_{0}^{8} e^{u} \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{8} e^{u} d u

Use the common integral:

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

= \frac{1}{4} [e^{u}]_{0}^{8}

Compute the boundaries:

e^{8} - 1

= \frac{1}{4} (e^{8} - 1)

Simplify

= \frac{e ^{8} - 1}{4}

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

x \to - 4 lim (3 ln (9) (7 - 5 x))

x \to 0 lim (sin (x) - e^{x})

\frac{d y}{d t} - 2 t y = - 12 t^{2} e^{t^{2}}, y (0) = - 1

\int (ln (4 x))^{2} d x

What is the integral from 0 to 2 of e^{4x} ?
The integral from 0 to 2 of e^{4x} is (e^8-1)/4