What is the integral from 0 to t of e^{-4.2804x} ?

The integral from 0 to t of e^{-4.2804x} is -0.23362…(e^{-4.2804t}-1)

Popular >

integral from 0 to t of e^{-4.2804x}

\int_{0}^{t} e^{- 4.2804 x} d x

- 0.23362 \dots (e^{- 4.2804 t} - 1)

Solution steps

\int_{0}^{t} e^{- 4.2804 x} d x

Apply u-substitution

: \int_{0}^{- 4.2804 t} - \frac{1}{4.2804} e^{u} d u

= \int_{0}^{- 4.2804 t} - \frac{1}{4.2804} e^{u} d u

Take the constant out:

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

= - \frac{1}{4.2804} \cdot \int_{0}^{- 4.2804 t} e^{u} d u

Simplify

= - 0.23362 \dots \cdot \int_{0}^{- 4.2804 t} e^{u} d u

Use the common integral:

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

= - 0.23362 \dots [e^{u}]_{0}^{- 4.2804 t}

Compute the boundaries:

e^{- 4.2804 t} - 1

= - 0.23362 \dots (e^{- 4.2804 t} - 1)

\int_{0}^{2 π} - 12 sin^{2} (t) d t

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

\int_{0}^{x} (t - 1)^{2} (t - 3)^{3} d t

\int_{0}^{0.2} \frac{x ^{6}}{1 + x ^{7}} d x

\int_{0}^{t} cos^{2} (x) d x

What is the integral from 0 to t of e^{-4.2804x} ?
The integral from 0 to t of e^{-4.2804x} is -0.23362…(e^{-4.2804t}-1)