What is the integral from 0 to 2 of 0.2e^{-0.2x} ?

The integral from 0 to 2 of 0.2e^{-0.2x} is 0.32967…

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integral from 0 to 2 of 0.2e^{-0.2x}

\int_{0}^{2} 0.2 e^{- 0.2 x} d x

0.32967 \dots

Solution steps

\int_{0}^{2} 0.2 e^{- 0.2 x} d x

Take the constant out:

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

= 0.2 \cdot \int_{0}^{2} e^{- 0.2 x} d x

Apply u-substitution

: \int_{0}^{- 0.4} - \frac{1}{0.2} e^{u} d u

= 0.2 \cdot \int_{0}^{- 0.4} - \frac{1}{0.2} e^{u} d u

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

= 0.2 (- \int_{- 0.4}^{0} - \frac{1}{0.2} e^{u} d u)

Take the constant out:

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

= 0.2 (- (- \frac{1}{0.2} \cdot \int_{- 0.4}^{0} e^{u} d u))

Simplify

= 0.2 (- (- 5 \cdot \int_{- 0.4}^{0} e^{u} d u))

Use the common integral:

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

= 0.2 (- (- 5 [e^{u}]_{- 0.4}^{0}))

Simplify

= [e^{u}]_{- 0.4}^{0}

Compute the boundaries:

0.32967 \dots

= 0.32967 \dots

y^{^{'}} - 5 y = t e^{(- 4 t)}

f (x) = \frac{1}{x ^{2}}, at x = 2

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{3}) = 0

h (31 - \frac{1}{2} h)^{2}

\int_{0}^{1} \frac{sin ( x )}{x} d x

What is the integral from 0 to 2 of 0.2e^{-0.2x} ?
The integral from 0 to 2 of 0.2e^{-0.2x} is 0.32967…