What is the integral from 3 to 5 of e^{-x/2} ?

The integral from 3 to 5 of e^{-x/2} is 2(1/(e^{3/2)}-1/(e^{5/2)})

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integral from 3 to 5 of e^{-x/2}

\int_{3}^{5} e^{- \frac{x}{2}} d x

2 (\frac{1}{e ^{\frac{3}{2}}} - \frac{1}{e ^{\frac{5}{2}}})

Decimal Notation

0.28209 \dots

Solution steps

\int_{3}^{5} e^{- \frac{x}{2}} d x

Apply u-substitution

: \int_{- \frac{3}{2}}^{- \frac{5}{2}} - 2 e^{u} d u

= \int_{- \frac{3}{2}}^{- \frac{5}{2}} - 2 e^{u} d u

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

= - \int_{- \frac{5}{2}}^{- \frac{3}{2}} - 2 e^{u} d u

Take the constant out:

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

= - (- 2 \cdot \int_{- \frac{5}{2}}^{- \frac{3}{2}} e^{u} d u)

Use the common integral:

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

= - (- 2 [e^{u}]_{- \frac{5}{2}}^{- \frac{3}{2}})

Simplify

= 2 [e^{u}]_{- \frac{5}{2}}^{- \frac{3}{2}}

Compute the boundaries:

\frac{1}{e ^{\frac{3}{2}}} - \frac{1}{e ^{\frac{5}{2}}}

= 2 (\frac{1}{e ^{\frac{3}{2}}} - \frac{1}{e ^{\frac{5}{2}}})

\int sin^{3} (16 x) d x

f (x) = \frac{1}{( x ^{3} + 7 ) ^{4}}

\frac{d}{d x} (2 x^{3} - 4 x)

\frac{2}{3 x ^{4}}

n = 1 \sum \infty \frac{n}{5 + n ^{2}}

What is the integral from 3 to 5 of e^{-x/2} ?
The integral from 3 to 5 of e^{-x/2} is 2(1/(e^{3/2)}-1/(e^{5/2)})