What is the integral from 0 to 2 of 4xe^x ?

The integral from 0 to 2 of 4xe^x is 4(e^2+1)

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integral from 0 to 2 of 4xe^x

\int_{0}^{2} 4 x e^{x} d x

4 (e^{2} + 1)

Decimal Notation

33.55622 \dots

Solution steps

\int_{0}^{2} 4 x e^{x} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{2} x e^{x} d x

Apply Integration By Parts

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

= 4 [e^{x} x - \int e^{x} d x]_{0}^{2}

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

= 4 [e^{x} x - e^{x}]_{0}^{2}

Compute the boundaries:

e^{2} + 1

= 4 (e^{2} + 1)

x \to 1 lim (\frac{4 x + 3}{2 x + 1})

n = 1 \sum \infty 4 \cdot (\frac{4}{7})^{n}

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f (x) = 5 x (sin (x) + cos (x))

\int sin (π + 2 y) - sin (2 y) d y

What is the integral from 0 to 2 of 4xe^x ?
The integral from 0 to 2 of 4xe^x is 4(e^2+1)