What is the integral from 0 to b of 1/32 ve^{-(v^2)/(64)} ?

The integral from 0 to b of 1/32 ve^{-(v^2)/(64)} is -e^{-(b^2)/(64)}+1

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integral from 0 to b of 1/32 ve^{-(v^2)/(64)}

\int_{0}^{b} \frac{1}{32} v e^{- \frac{v ^{2}}{64}} d v

- e^{- \frac{b ^{2}}{64}} + 1

Solution steps

\int_{0}^{b} \frac{1}{32} v e^{- \frac{v ^{2}}{64}} d v

Take the constant out:

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

= \frac{1}{32} \cdot \int_{0}^{b} v e^{- \frac{v ^{2}}{64}} d v

Apply u-substitution

: \int_{0}^{- \frac{b ^{2}}{64}} - 32 e^{u} d u

= \frac{1}{32} \cdot \int_{0}^{- \frac{b ^{2}}{64}} - 32 e^{u} d u

Take the constant out:

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

= \frac{1}{32} (- 32 \cdot \int_{0}^{- \frac{b ^{2}}{64}} e^{u} d u)

Use the common integral:

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

= \frac{1}{32} (- 32 [e^{u}]_{0}^{- \frac{b ^{2}}{64}})

Simplify

\frac{1}{32} (- 32 [e^{u}]_{0}^{- \frac{b ^{2}}{64}}) : - [e^{u}]_{0}^{- \frac{b ^{2}}{64}}

= - [e^{u}]_{0}^{- \frac{b ^{2}}{64}}

Compute the boundaries:

e^{- \frac{b ^{2}}{64}} - 1

= - (e^{- \frac{b ^{2}}{64}} - 1)

Simplify

= - e^{- \frac{b ^{2}}{64}} + 1

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What is the integral from 0 to b of 1/32 ve^{-(v^2)/(64)} ?
The integral from 0 to b of 1/32 ve^{-(v^2)/(64)} is -e^{-(b^2)/(64)}+1