What is the integral of 32e^{-4}x ?

The integral of 32e^{-4}x is (16x^2)/(e^4)+C

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integral of 32e^{-4}x

integral

32 e^{- 4} x

\frac{16 x ^{2}}{e ^{4}} + C

Solution steps

\int 32 e^{- 4} x d x

Take the constant out:

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

= 32 e^{- 4} \cdot \int x d x

32 e^{- 4} = \frac{32}{e ^{4}}

= \frac{32}{e ^{4}} \cdot \int x d x

Apply the Power Rule

: \frac{x ^{2}}{2}

= \frac{32}{e ^{4}} \cdot \frac{x ^{2}}{2}

Simplify

\frac{32}{e ^{4}} \cdot \frac{x ^{2}}{2} : \frac{16 x ^{2}}{e ^{4}}

= \frac{16 x ^{2}}{e ^{4}}

Add a constant to the solution

= \frac{16 x ^{2}}{e ^{4}} + C

\frac{( x + 3 )}{( 2 x + 1 )}

5 x^{2} - 2

x^{2} e^{x^{2}}

2 x sin (x^{2} + 1)

\frac{( x + 3 )}{( x ^{2} - x + 6 )}

What is the integral of 32e^{-4}x ?
The integral of 32e^{-4}x is (16x^2)/(e^4)+C