What is the integral from 2 to 4 of e^{3x-y} ?

The integral from 2 to 4 of e^{3x-y} is e^{3x}(1/(e^2)-1/(e^4))

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integral from 2 to 4 of e^{3x-y}

\int_{2}^{4} e^{3 x - y} d y

e^{3 x} (\frac{1}{e ^{2}} - \frac{1}{e ^{4}})

Solution steps

\int_{2}^{4} e^{3 x - y} d y

e^{- y + 3 x} = e^{3 x} e^{- y}

= \int_{2}^{4} e^{3 x} e^{- y} d y

Take the constant out:

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

= e^{3 x} \cdot \int_{2}^{4} e^{- y} d y

Apply u-substitution

: \int_{- 2}^{- 4} - e^{u} d u

= e^{3 x} \cdot \int_{- 2}^{- 4} - e^{u} d u

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

= e^{3 x} (- \int_{- 4}^{- 2} - e^{u} d u)

Take the constant out:

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

= e^{3 x} (- (- \int_{- 4}^{- 2} e^{u} d u))

Use the common integral:

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

= e^{3 x} (- (- [e^{u}]_{- 4}^{- 2}))

Simplify

= e^{3 x} [e^{u}]_{- 4}^{- 2}

Compute the boundaries:

\frac{1}{e ^{2}} - \frac{1}{e ^{4}}

= e^{3 x} (\frac{1}{e ^{2}} - \frac{1}{e ^{4}})

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

x \to 0 lim (\frac{5 x ^{2}}{2})

y^{^{'}} + \frac{1}{t} y = \frac{1}{t ^{2}}

f (x) = x^{2} cos (x)

\int e^{(- \frac{x}{4})} d x

What is the integral from 2 to 4 of e^{3x-y} ?
The integral from 2 to 4 of e^{3x-y} is e^{3x}(1/(e^2)-1/(e^4))