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

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

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integral from 0 to 2 of e^{2t}

\int_{0}^{2} e^{2 t} d t

\frac{e ^{4} - 1}{2}

Decimal Notation

26.79907 \dots

Solution steps

\int_{0}^{2} e^{2 t} d t

Apply u-substitution

: \int_{0}^{4} e^{u} \frac{1}{2} d u

= \int_{0}^{4} e^{u} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= \frac{1}{2} (e^{4} - 1)

Simplify

= \frac{e ^{4} - 1}{2}

3 - x^{2}, - 1

y^{^{''}} + 4 y^{^{'}} + 25 y = 0

n = 1 \sum \infty \frac{5 n}{n ^{2} + 4}

y = x^{2} - 3 x - 5, (2, - 7)

\frac{d}{d x} (\frac{6000}{x})

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