What is the integral from 0 to 0.5 of-1/10 e^{-10y} ?

The integral from 0 to 0.5 of-1/10 e^{-10y} is -(e^5-1)/(100e^5)

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integral from 0 to 0.5 of-1/10 e^{-10y}

\int_{0}^{0.5} - \frac{1}{10} e^{- 10 y} d y

- \frac{e ^{5} - 1}{100 e ^{5}}

Decimal

- 0.00993 \dots

Solution steps

\int_{0}^{0.5} - \frac{1}{10} e^{- 10 y} d y

Take the constant out:

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

= - \frac{1}{10} \cdot \int_{0}^{0.5} e^{- 10 y} d y

Apply u-substitution

: \int_{0}^{- 5} - \frac{1}{10} e^{u} d u

= - \frac{1}{10} \cdot \int_{0}^{- 5} - \frac{1}{10} e^{u} d u

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

= - \frac{1}{10} (- \int_{- 5}^{0} - \frac{1}{10} e^{u} d u)

Take the constant out:

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

= - \frac{1}{10} (- (- \frac{1}{10} \cdot \int_{- 5}^{0} e^{u} d u))

Use the common integral:

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

= - \frac{1}{10} (- (- \frac{1}{10} [e^{u}]_{- 5}^{0}))

Simplify

- \frac{1}{10} (- (- \frac{1}{10} [e^{u}]_{- 5}^{0})) : - \frac{1}{100} [e^{u}]_{- 5}^{0}

= - \frac{1}{100} [e^{u}]_{- 5}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{5}}

= - \frac{1}{100} (1 - \frac{1}{e ^{5}})

Simplify

= - \frac{e ^{5} - 1}{100 e ^{5}}

lo g_{100} (x), (0, 24)

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f (x) = \frac{x ^{2} e ^{x}}{x ^{2} + e ^{x}}

\int \frac{1}{x} \cdot x d x

What is the integral from 0 to 0.5 of-1/10 e^{-10y} ?
The integral from 0 to 0.5 of-1/10 e^{-10y} is -(e^5-1)/(100e^5)