What is the integral from 0 to 3 of 1/4 e^{-x/4} ?

The integral from 0 to 3 of 1/4 e^{-x/4} is 1-1/(e^{3/4)}

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integral from 0 to 3 of 1/4 e^{-x/4}

\int_{0}^{3} \frac{1}{4} e^{- \frac{x}{4}} d x

1 - \frac{1}{e ^{\frac{3}{4}}}

Decimal

0.52763 \dots

Solution steps

\int_{0}^{3} \frac{1}{4} e^{- \frac{x}{4}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{3} e^{- \frac{x}{4}} d x

Apply u-substitution

= \frac{1}{4} \cdot \int_{0}^{- \frac{3}{4}} - 4 e^{u} d u

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

= \frac{1}{4} (- \int_{- \frac{3}{4}}^{0} - 4 e^{u} d u)

Take the constant out:

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

= \frac{1}{4} (- (- 4 \cdot \int_{- \frac{3}{4}}^{0} e^{u} d u))

Use the common integral:

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

= \frac{1}{4} (- (- 4 [e^{u}]_{- \frac{3}{4}}^{0}))

Simplify

\frac{1}{4} (- (- 4 [e^{u}]_{- \frac{3}{4}}^{0})) : [e^{u}]_{- \frac{3}{4}}^{0}

= [e^{u}]_{- \frac{3}{4}}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{\frac{3}{4}}}

= 1 - \frac{1}{e ^{\frac{3}{4}}}

What is the integral from 0 to 3 of 1/4 e^{-x/4} ?
The integral from 0 to 3 of 1/4 e^{-x/4} is 1-1/(e^{3/4)}