What is the integral of (1/8)y_{1}e^{-(y_{1}+y_{2})/2}_{2} ?

The integral of (1/8)y_{1}e^{-(y_{1}+y_{2})/2}_{2} is -e^{-1/2 (y_{1}+y_{2})}(y_{1})/4+C

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integral of (1/8)y_{1}e^{-(y_{1}+y_{2})/2}_{2}

\int (\frac{1}{8}) y_{1} e^{- \frac{y _{1} + y _{2}}{2}} d y_{2}

- e^{- \frac{1}{2} (y_{1} + y_{2})} \frac{y _{1}}{4} + C

Solution steps

\int \frac{1}{8} y_{1} e^{- \frac{y _{1} + y _{2}}{2}} d y_{2}

Take the constant out:

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

= \frac{1}{8} y_{1} \cdot \int e^{- \frac{y _{1} + y _{2}}{2}} d y_{2}

Simplify

\frac{1}{8} y_{1} : \frac{y _{1}}{8}

= \frac{y _{1}}{8} \cdot \int e^{- \frac{y _{1} + y _{2}}{2}} d y_{2}

Apply u-substitution

: \int e^{- \frac{u}{2}} d u

= \frac{y _{1}}{8} \cdot \int e^{- \frac{u}{2}} d u

Apply u-substitution

: \int - 2 e^{v} d v

= \frac{y _{1}}{8} \cdot \int - 2 e^{v} d v

Take the constant out:

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

= \frac{y _{1}}{8} (- 2 \cdot \int e^{v} d v)

Use the common integral:

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

= \frac{y _{1}}{8} (- 2 e^{v})

Substitute back

= \frac{y _{1}}{8} (- 2 e^{- \frac{y _{1} + y _{2}}{2}})

Simplify

\frac{y _{1}}{8} (- 2 e^{- \frac{y _{1} + y _{2}}{2}}) : - e^{- \frac{1}{2} (y_{1} + y_{2})} \frac{y _{1}}{4}

= - e^{- \frac{1}{2} (y_{1} + y_{2})} \frac{y _{1}}{4}

Add a constant to the solution

= - e^{- \frac{1}{2} (y_{1} + y_{2})} \frac{y _{1}}{4} + C

f (x) = 3 x^{2}

x \to - 2 lim (x^{2} + 2)

x \to 1 + lim (\frac{( x - 5 )}{x - 1})

\frac{d}{d x} ((4 x^{3} + 5 x)^{4})

f (x) = x ln (x)

What is the integral of (1/8)y_{1}e^{-(y_{1}+y_{2})/2}_{2} ?
The integral of (1/8)y_{1}e^{-(y_{1}+y_{2})/2}_{2} is -e^{-1/2 (y_{1}+y_{2})}(y_{1})/4+C