What is the integral from-1 to 3 of e^{u+1} ?

The integral from-1 to 3 of e^{u+1} is e^4-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 3 of e^{u+1}

\int_{- 1}^{3} e^{u + 1} d u

e^{4} - 1

Decimal

53.59815 \dots

Solution steps

\int_{- 1}^{3} e^{u + 1} d u

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

= \int_{- 1}^{3} 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}^{3} e^{u} d u

Simplify

= e \cdot \int_{- 1}^{3} e^{u} d u

Use the common integral:

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

= e [e^{u}]_{- 1}^{3}

Compute the boundaries:

e^{3} - \frac{1}{e}

= e (e^{3} - \frac{1}{e})

Simplify

= e^{4} - 1

\int_{7}^{8} - 7 lo g_{2} (t) d t

\int_{1}^{5} x^{5} ln (x) d x

\int_{0}^{3} - 2 x 9 - x^{2} d x

\int_{- 1}^{\infty} \frac{1}{x} d x

\int_{0}^{4} (x^{3} + 2) d x

What is the integral from-1 to 3 of e^{u+1} ?
The integral from-1 to 3 of e^{u+1} is e^4-1