What is the integral from-2 to 0 of 2e^{4x-1} ?

The integral from-2 to 0 of 2e^{4x-1} is (e^8-1)/(2e^9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-2 to 0 of 2e^{4x-1}

\int_{- 2}^{0} 2 e^{4 x - 1} d x

\frac{e ^{8} - 1}{2 e ^{9}}

Decimal Notation

0.18387 \dots

Solution steps

\int_{- 2}^{0} 2 e^{4 x - 1} d x

Take the constant out:

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

= 2 \cdot \int_{- 2}^{0} e^{4 x - 1} d x

Apply u-substitution

: \int_{- 9}^{- 1} e^{u} \frac{1}{4} d u

= 2 \cdot \int_{- 9}^{- 1} e^{u} \frac{1}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int_{- 9}^{- 1} e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{4} [e^{u}]_{- 9}^{- 1}

Simplify

2 \cdot \frac{1}{4} [e^{u}]_{- 9}^{- 1} : \frac{1}{2} [e^{u}]_{- 9}^{- 1}

= \frac{1}{2} [e^{u}]_{- 9}^{- 1}

Compute the boundaries:

\frac{1}{e} - \frac{1}{e ^{9}}

= \frac{1}{2} (\frac{1}{e} - \frac{1}{e ^{9}})

Simplify

= \frac{e ^{8} - 1}{2 e ^{9}}

\int_{4}^{9} 7 x d x

\int_{0}^{x + y} (x + y + z) d z

\int_{- \frac{π}{2}}^{\frac{π}{2}} 5 cos (x) d x

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

\int_{1}^{\infty} \frac{4}{x ^{5}} d x

What is the integral from-2 to 0 of 2e^{4x-1} ?
The integral from-2 to 0 of 2e^{4x-1} is (e^8-1)/(2e^9)