What is the integral of 10e^{-10x} ?

The integral of 10e^{-10x} is -e^{-10x}+C

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integral of 10e^{-10x}

\int 10 e^{- 10 x} d x

- e^{- 10 x} + C

Solution steps

\int 10 e^{- 10 x} d x

Take the constant out:

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

= 10 \cdot \int e^{- 10 x} d x

Apply u-substitution

: \int - \frac{1}{10} e^{u} d u

= 10 \cdot \int - \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

= 10 (- \frac{1}{10} \cdot \int e^{u} d u)

Use the common integral:

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

= 10 (- \frac{1}{10} e^{u})

Substitute back

u = - 10 x

= 10 (- \frac{1}{10} e^{- 10 x})

Simplify

10 (- \frac{1}{10} e^{- 10 x}) : - e^{- 10 x}

= - e^{- 10 x}

Add a constant to the solution

= - e^{- 10 x} + C

\int - \frac{8}{π} x d x

f (x) = \frac{2}{1 + e ^{- x}}, (0, 1)

x^{^{'}} = (x - 17)^{2}

f (x) = \frac{7}{x + 8}, (5, \frac{7}{13})

y^{^{'}} = 0.06 (y - 65)