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

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

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

e^{4} - 1

Decimal

53.59815 \dots

Solution steps

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

e^{x + 1} = e^{1} e^{x}

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

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^{x} d x

Simplify

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

Use the common integral:

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

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

Compute the boundaries:

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

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

Simplify

= e^{4} - 1

\int_{0}^{1} x 24 x^{2} + 25 d x

\int_{0}^{1} \frac{x ^{3}}{3} d x

\int_{1}^{7} \frac{3}{x} d x

\int_{2}^{3} (x^{3} - \frac{1}{x}) d x

\int_{π}^{l n (x)} cos (e^{t}) d t

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