What is the integral from 0 to 1 of 7e^{-2x} ?

The integral from 0 to 1 of 7e^{-2x} is (7(e^2-1))/(2e^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 7e^{-2x}

\int_{0}^{1} 7 e^{- 2 x} d x

\frac{7 ( e ^{2} - 1 )}{2 e ^{2}}

Decimal

3.02632 \dots

Solution steps

\int_{0}^{1} 7 e^{- 2 x} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 2} - \frac{1}{2} e^{u} d u

= 7 \cdot \int_{0}^{- 2} - \frac{1}{2} e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 7 (- \int_{- 2}^{0} - \frac{1}{2} e^{u} d u)

Take the constant out:

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

= 7 (- (- \frac{1}{2} \cdot \int_{- 2}^{0} e^{u} d u))

Use the common integral:

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

= 7 (- (- \frac{1}{2} [e^{u}]_{- 2}^{0}))

Simplify

7 (- (- \frac{1}{2} [e^{u}]_{- 2}^{0})) : \frac{7}{2} [e^{u}]_{- 2}^{0}

= \frac{7}{2} [e^{u}]_{- 2}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{2}}

= \frac{7}{2} (1 - \frac{1}{e ^{2}})

Simplify

= \frac{7 ( e ^{2} - 1 )}{2 e ^{2}}

x \to - \infty lim (1^{x})

f (x) = 4 - 3 x^{4} - 2

4 cos (x) + 12 x

\int_{1}^{2} \frac{5}{x ^{2}} d x

f (x) = 8 x - 3, at x = 64

What is the integral from 0 to 1 of 7e^{-2x} ?
The integral from 0 to 1 of 7e^{-2x} is (7(e^2-1))/(2e^2)