What is the integral from 1 to 2 of (e^{(x-1)})^2 ?

The integral from 1 to 2 of (e^{(x-1)})^2 is (e^4-e^2)/(2e^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of (e^{(x-1)})^2

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

\frac{e ^{4} - e ^{2}}{2 e ^{2}}

Decimal Notation

3.19452 \dots

Solution steps

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

(e^{x - 1})^{2} = (e^{- 1})^{2} (e^{x})^{2}

= \int_{1}^{2} (e^{- 1})^{2} (e^{x})^{2} d x

Take the constant out:

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

= (e^{- 1})^{2} \cdot \int_{1}^{2} (e^{x})^{2} d x

(e^{- 1})^{2} = \frac{1}{e ^{2}}

= \frac{1}{e ^{2}} \cdot \int_{1}^{2} (e^{x})^{2} d x

Apply u-substitution

: \int_{e}^{e^{2}} u d u

= \frac{1}{e ^{2}} \cdot \int_{e}^{e^{2}} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{e}^{e^{2}}

= \frac{1}{e ^{2}} [\frac{u ^{2}}{2}]_{e}^{e^{2}}

Compute the boundaries:

\frac{e ^{4} - e ^{2}}{2}

= \frac{1}{e ^{2}} \cdot \frac{e ^{4} - e ^{2}}{2}

Simplify

= \frac{e ^{4} - e ^{2}}{2 e ^{2}}

\int_{5}^{5} 3 x^{2} (8 - x^{3})^{4} d x

\int_{0}^{7} π sin (\frac{π t}{7}) + 1 d t

\int_{2}^{6} 4 x + 4 d x

\int_{0}^{5} 5 x^{2} - x^{3} d x

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

What is the integral from 1 to 2 of (e^{(x-1)})^2 ?
The integral from 1 to 2 of (e^{(x-1)})^2 is (e^4-e^2)/(2e^2)