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

The integral from 0 to 4 of xe^{x+2} is 3e^6+e^2

Popular >

integral from 0 to 4 of xe^{x+2}

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

3 e^{6} + e^{2}

Solution steps

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

x e^{x + 2} = e^{2} x e^{x}

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

3 e^{4} + 1

= e^{2} (3 e^{4} + 1)

Simplify

= 3 e^{6} + e^{2}

\int \frac{5 - 3 sin ( 2 x )}{sec ( 6 x )} d x

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

\frac{1}{4 + x ^{2}}

\int_{0}^{\frac{π}{2}} (sin^{3} (x)) d x

n = 2 \sum \infty \frac{n}{3 ^{n + 1}}

What is the integral from 0 to 4 of xe^{x+2} ?
The integral from 0 to 4 of xe^{x+2} is 3e^6+e^2