What is the integral of 2000(x-4)e^{x^2-8x} ?

The integral of 2000(x-4)e^{x^2-8x} is 1000e^{x^2-8x}+C

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integral of 2000(x-4)e^{x^2-8x}

\int 2000 (x - 4) e^{x^{2} - 8 x} d x

1000 e^{x^{2} - 8 x} + C

Solution steps

\int 2000 (x - 4) e^{x^{2} - 8 x} d x

Take the constant out:

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

= 2000 \cdot \int (x - 4) e^{x^{2} - 8 x} d x

Apply u-substitution

: \int e^{u^{2} - 16} u d u

= 2000 \cdot \int e^{u^{2} - 16} u d u

Apply u-substitution

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

= 2000 \cdot \int \frac{e ^{v}}{2} d v

Take the constant out:

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

= 2000 \cdot \frac{1}{2} \cdot \int e^{v} d v

Use the common integral:

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

= 2000 \cdot \frac{1}{2} e^{v}

Substitute back

= 2000 \cdot \frac{1}{2} e^{(x - 4)^{2} - 16}

Simplify

2000 \cdot \frac{1}{2} e^{(x - 4)^{2} - 16} : 1000 e^{x^{2} - 8 x}

= 1000 e^{x^{2} - 8 x}

Add a constant to the solution

= 1000 e^{x^{2} - 8 x} + C

2 x^{2} - 4

f (x) = x^{3} + 2

n = 1 \sum \infty 1 7^{n} x^{n} n!

\frac{d y}{d x} = 3 x^{2} (y^{2} + 1)

\frac{d}{d x} (x - 10)

What is the integral of 2000(x-4)e^{x^2-8x} ?
The integral of 2000(x-4)e^{x^2-8x} is 1000e^{x^2-8x}+C