What is the integral from 0 to 4 of pie^{-x/2} ?

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

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integral from 0 to 4 of pie^{-x/2}

\int_{0}^{4} π e^{- \frac{x}{2}} d x

2 π (1 - \frac{1}{e ^{2}})

Decimal

5.43284 \dots

Solution steps

\int_{0}^{4} π e^{- \frac{x}{2}} d x

Take the constant out:

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

= π \cdot \int_{0}^{4} e^{- \frac{x}{2}} d x

Apply u-substitution

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

= π \cdot \int_{0}^{- 2} - 2 e^{u} d u

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

= π (- \int_{- 2}^{0} - 2 e^{u} d u)

Take the constant out:

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

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

Use the common integral:

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

= π (- (- 2 [e^{u}]_{- 2}^{0}))

Simplify

= 2 π [e^{u}]_{- 2}^{0}

Compute the boundaries:

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

= 2 π (1 - \frac{1}{e ^{2}})

x \to 0 lim (2^{(x - 1)})

\frac{4}{s + 5}

\frac{d}{d x} (- 4 e^{x} sin (x))

\int \frac{1}{x ^{4} 16 - 25 x ^{2}} d x

\frac{d}{d x} (\frac{x ^{2} - 16}{x + 4})

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