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

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

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integral from 0 to 4 of e^{-2t}

\int_{0}^{4} e^{- 2 t} d t

\frac{e ^{8} - 1}{2 e ^{8}}

Decimal

0.49983 \dots

Solution steps

\int_{0}^{4} e^{- 2 t} d t

Apply u-substitution

: \int_{0}^{- 8} - \frac{1}{2} e^{u} d u

= \int_{0}^{- 8} - \frac{1}{2} e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 8}^{0} - \frac{1}{2} e^{u} 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_{- 8}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{- 8}^{0})

Simplify

= \frac{1}{2} [e^{u}]_{- 8}^{0}

Compute the boundaries:

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

= \frac{1}{2} (1 - \frac{1}{e ^{8}})

Simplify

= \frac{e ^{8} - 1}{2 e ^{8}}

\int_{1}^{3} 2 x^{2} + 1 d x

\int_{- 3}^{3} (x^{3} + 3 x^{2} + 2 x) d x

\int_{1}^{3} 2 x^{2} + 4 d x

\int_{0}^{\frac{π}{2}} x sin (y) d y

\int_{- 1}^{0} 2 y + 2 d y

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