What is the integral from 0 to 0.2 of 8e^{-8x} ?

The integral from 0 to 0.2 of 8e^{-8x} is 0.79810…

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integral from 0 to 0.2 of 8e^{-8x}

\int_{0}^{0.2} 8 e^{- 8 x} d x

0.79810 \dots

Solution steps

\int_{0}^{0.2} 8 e^{- 8 x} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{0.2} e^{- 8 x} d x

Apply u-substitution

: \int_{0}^{- 1.6} - \frac{1}{8} e^{u} d u

= 8 \cdot \int_{0}^{- 1.6} - \frac{1}{8} e^{u} d u

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

= 8 (- \int_{- 1.6}^{0} - \frac{1}{8} e^{u} d u)

Take the constant out:

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

= 8 (- (- \frac{1}{8} \cdot \int_{- 1.6}^{0} e^{u} d u))

Use the common integral:

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

= 8 (- (- \frac{1}{8} [e^{u}]_{- 1.6}^{0}))

Simplify

8 (- (- \frac{1}{8} [e^{u}]_{- 1.6}^{0})) : [e^{u}]_{- 1.6}^{0}

= [e^{u}]_{- 1.6}^{0}

Compute the boundaries:

0.79810 \dots

= 0.79810 \dots

\int x (e^{x} + e^{x^{2}}) d x

2 x^{2} + 11, 0, - 2, 2

f (x) = x^{3} - 2 x + 2, at x = 4

\frac{d}{d x} (ln (x + ln (x)))

\frac{d}{d x} (1 - e^{- \frac{x}{2}})

What is the integral from 0 to 0.2 of 8e^{-8x} ?
The integral from 0 to 0.2 of 8e^{-8x} is 0.79810…