What is the integral from 0 to 1 of 34.6e^{-0.06t} ?

The integral from 0 to 1 of 34.6e^{-0.06t} is 33.58245…

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integral from 0 to 1 of 34.6e^{-0.06t}

\int_{0}^{1} 34.6 e^{- 0.06 t} d t

33.58245 \dots

Solution steps

\int_{0}^{1} 34.6 e^{- 0.06 t} d t

Take the constant out:

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

= 34.6 \cdot \int_{0}^{1} e^{- 0.06 t} d t

Apply u-substitution

: \int_{0}^{- 0.06} - \frac{1}{0.06} e^{u} d u

= 34.6 \cdot \int_{0}^{- 0.06} - \frac{1}{0.06} e^{u} d u

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

= 34.6 (- \int_{- 0.06}^{0} - \frac{1}{0.06} e^{u} d u)

Take the constant out:

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

= 34.6 (- (- \frac{1}{0.06} \cdot \int_{- 0.06}^{0} e^{u} d u))

Simplify

= 34.6 (- (- 16.66666 \dots \cdot \int_{- 0.06}^{0} e^{u} d u))

Use the common integral:

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

= 34.6 (- (- 16.66666 \dots [e^{u}]_{- 0.06}^{0}))

Simplify

= 576.66666 \dots [e^{u}]_{- 0.06}^{0}

Compute the boundaries:

0.05823 \dots

= 576.66666 \dots \cdot 0.05823 \dots

Simplify

= 33.58245 \dots

\int_{0}^{0.15} x d x

x \to 0 lim (x^{9} cos (\frac{4}{x}))

\int \frac{1}{x 49 x ^{4} - 4} d x

y^{^{''}} - y^{^{'}} - 2 y = t e^{2 t}

- \frac{1}{9} (x^{- 9} - x^{18})

What is the integral from 0 to 1 of 34.6e^{-0.06t} ?
The integral from 0 to 1 of 34.6e^{-0.06t} is 33.58245…