What is the integral of (-3+4)e^{-x} ?

The integral of (-3+4)e^{-x} is -e^{-x}+C

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integral of (-3+4)e^{-x}

\int (- 3 + 4) e^{- x} d x

- e^{- x} + C

Solution steps

\int (- 3 + 4) e^{- x} d x

Take the constant out:

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

= (- 3 + 4) \cdot \int e^{- x} d x

Simplify

= 1 \cdot \int e^{- x} d x

Apply u-substitution

= 1 \cdot \int - e^{u} d u

Take the constant out:

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

= 1 \cdot (- \int e^{u} d u)

Use the common integral:

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

= 1 \cdot (- e^{u})

Substitute back

u = - x

= 1 \cdot (- e^{- x})

Simplify

= - e^{- x}

Add a constant to the solution

= - e^{- x} + C

(1 + x) \frac{d y}{d x} = 4 y

\frac{d}{d x} (2 x^{0.5})

\int 2 sin^{3} (x co) s^{7} x

t an g e n t f (x) = \frac{9}{x}, at x = 5

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

What is the integral of (-3+4)e^{-x} ?
The integral of (-3+4)e^{-x} is -e^{-x}+C