What is the integral from 0 to 1 of 2xe^x ?

The integral from 0 to 1 of 2xe^x is 2

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integral from 0 to 1 of 2xe^x

\int_{0}^{1} 2 x e^{x} d x

2

Solution steps

\int_{0}^{1} 2 x e^{x} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} x e^{x} d x

Apply Integration By Parts

= 2 [e^{x} x - \int e^{x} d x]_{0}^{1}

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

= 2 [e^{x} x - e^{x}]_{0}^{1}

Compute the boundaries:

1

= 2 \cdot 1

Simplify

= 2

\int \frac{cos ^{2} ( x )}{1 + sin ( x )} d x

f (x) = ln (x + 3)

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

\frac{d}{d y} (\frac{e ^{y} + e ^{- y}}{2})

\frac{d}{d x} (\frac{1}{3} sin^{3} (x) - sin (x) + 1)

What is the integral from 0 to 1 of 2xe^x ?
The integral from 0 to 1 of 2xe^x is 2