What is the integral from 0 to 8 of e^{-0.25t} ?

The integral from 0 to 8 of e^{-0.25t} is 4(1-1/(e^2))

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integral from 0 to 8 of e^{-0.25t}

\int_{0}^{8} e^{- 0.25 t} d t

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

Decimal

3.45865 \dots

Solution steps

\int_{0}^{8} e^{- 0.25 t} d t

Apply u-substitution

: \int_{0}^{- 2} - \frac{1}{0.25} e^{u} d u

= \int_{0}^{- 2} - \frac{1}{0.25} e^{u} d u

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

= - \int_{- 2}^{0} - \frac{1}{0.25} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{0.25} \cdot \int_{- 2}^{0} e^{u} d u)

Simplify

= - (- 4 \cdot \int_{- 2}^{0} e^{u} d u)

Use the common integral:

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

= - (- 4 [e^{u}]_{- 2}^{0})

Simplify

= 4 [e^{u}]_{- 2}^{0}

Compute the boundaries:

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

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

\int sin (3 x) sin (6 x) d x

[x^{2} (x^{2} + 8 x)]^{6}

x^{2} \frac{d y}{d x} = y^{2}

\frac{x ^{4}}{16} + \frac{1}{2 x ^{2}}

\int \frac{1}{16 - 4 x ^{2}} d x

What is the integral from 0 to 8 of e^{-0.25t} ?
The integral from 0 to 8 of e^{-0.25t} is 4(1-1/(e^2))