What is the integral from-1 to 1 of e^{u+1} ?

The integral from-1 to 1 of e^{u+1} is e^2-1

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integral from-1 to 1 of e^{u+1}

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

e^{2} - 1

Decimal

6.38905 \dots

Solution steps

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

e^{u + 1} = e^{1} e^{u}

= \int_{- 1}^{1} e^{1} e^{u} d u

Take the constant out:

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

= e^{1} \cdot \int_{- 1}^{1} e^{u} d u

Simplify

= e \cdot \int_{- 1}^{1} e^{u} d u

Use the common integral:

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

= e [e^{u}]_{- 1}^{1}

Compute the boundaries:

e - \frac{1}{e}

= e (e - \frac{1}{e})

Simplify

= e^{2} - 1

x \to 2 lim (\frac{x ^{2} + 1}{x - 2})

\frac{\partial}{\partial x} (x - \frac{2 y + 6 x}{13})

y = ln (\frac{x}{1 + x ^{2}})

\frac{4}{10} x^{2} d y - 13 d x = 0

\frac{3}{( s - 2 ) ^{2}}

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