What is the integral from 0 to 1 of (x-3)e^x ?

The integral from 0 to 1 of (x-3)e^x is -3e+4

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integral from 0 to 1 of (x-3)e^x

\int_{0}^{1} (x - 3) e^{x} d x

- 3 e + 4

Decimal

- 4.15484 \dots

Solution steps

\int_{0}^{1} (x - 3) e^{x} d x

Apply Integration By Parts

: [e^{x} (x - 3) - \int e^{x} d x]_{0}^{1}

= [e^{x} (x - 3) - \int e^{x} d x]_{0}^{1}

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

= [e^{x} (x - 3) - e^{x}]_{0}^{1}

Expand

e^{x} (x - 3) - e^{x} : e^{x} x - 4 e^{x}

= [e^{x} x - 4 e^{x}]_{0}^{1}

Compute the boundaries:

- 3 e + 4

= - 3 e + 4

f (x) = 8 x^{5} - 8 cos (x)

\frac{d}{d x} (x^{x c o s (x)})

x^{2} + 4 ln^{2} (x)

ln (x^{2})

\frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}}

What is the integral from 0 to 1 of (x-3)e^x ?
The integral from 0 to 1 of (x-3)e^x is -3e+4