What is the integral from 0 to 9 of pie^{-(2x)/9} ?

The integral from 0 to 9 of pie^{-(2x)/9} is (9pi(e^2-1))/(2e^2)

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integral from 0 to 9 of pie^{-(2x)/9}

\int_{0}^{9} π e^{- \frac{2 x}{9}} d x

\frac{9 π ( e ^{2} - 1 )}{2 e ^{2}}

Decimal

12.22390 \dots

Solution steps

\int_{0}^{9} π e^{- \frac{2 x}{9}} d x

Take the constant out:

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

= π \cdot \int_{0}^{9} e^{- \frac{2 x}{9}} d x

Apply u-substitution

= π \cdot \int_{0}^{- 2} - \frac{9}{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} - \frac{9}{2} e^{u} d u)

Take the constant out:

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

= π (- (- \frac{9}{2} \cdot \int_{- 2}^{0} e^{u} d u))

Use the common integral:

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

= π (- (- \frac{9}{2} [e^{u}]_{- 2}^{0}))

Simplify

= π \frac{9}{2} [e^{u}]_{- 2}^{0}

Compute the boundaries:

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

= π \frac{9}{2} (1 - \frac{1}{e ^{2}})

Simplify

= \frac{9 π ( e ^{2} - 1 )}{2 e ^{2}}

\int x^{3} x^{2} + 11 d x

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{x}

\int_{0}^{5} \frac{1}{25 - x ^{2}} d x

x \to - 1 - lim (\frac{9}{x ^{3} - 1})

\int \frac{3}{16 - x ^{2}} d x

What is the integral from 0 to 9 of pie^{-(2x)/9} ?
The integral from 0 to 9 of pie^{-(2x)/9} is (9pi(e^2-1))/(2e^2)