What is the integral from-1 to 1 of xe^{x+1} ?

The integral from-1 to 1 of xe^{x+1} is 2

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integral from-1 to 1 of xe^{x+1}

\int_{- 1}^{1} x e^{x + 1} d x

2

Solution steps

\int_{- 1}^{1} x e^{x + 1} d x

Apply Integration By Parts

: [e^{1 + x} x - \int e^{1 + x} d x]_{- 1}^{1}

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

\int e^{1 + x} d x = e^{1 + x}

= [e^{1 + x} x - e^{1 + x}]_{- 1}^{1}

Compute the boundaries:

2

= 2

x \to 0 lim (9 x - 4)

\int \frac{1}{16 + 25 x ^{2}} + \frac{2}{3 ( x - 2 )} d x

x \to π lim (- 9 sin (\frac{x}{2}))

f (x) = (- 9 x^{2} + 16)^{2}

\int 6 x d x

What is the integral from-1 to 1 of xe^{x+1} ?
The integral from-1 to 1 of xe^{x+1} is 2