What is the integral from 1 to 2 of 3/(e^{4t)} ?

The integral from 1 to 2 of 3/(e^{4t)} is (3(e^4-1))/(4e^8)

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 3/(e^{4t)}

\int_{1}^{2} \frac{3}{e ^{4 t}} d t

\frac{3 ( e ^{4} - 1 )}{4 e ^{8}}

Decimal Notation

0.01348 \dots

Solution steps

\int_{1}^{2} \frac{3}{e ^{4 t}} d t

Take the constant out:

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

= 3 \cdot \int_{1}^{2} \frac{1}{e ^{4 t}} d t

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{4 t}} = e^{- 4 t}

= 3 \cdot \int_{1}^{2} e^{- 4 t} d t

Apply u-substitution

: \int_{- 4}^{- 8} - \frac{1}{4} e^{u} d u

= 3 \cdot \int_{- 4}^{- 8} - \frac{1}{4} e^{u} d u

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

= 3 (- \int_{- 8}^{- 4} - \frac{1}{4} e^{u} d u)

Take the constant out:

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

= 3 (- (- \frac{1}{4} \cdot \int_{- 8}^{- 4} e^{u} d u))

Use the common integral:

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

= 3 (- (- \frac{1}{4} [e^{u}]_{- 8}^{- 4}))

Simplify

3 (- (- \frac{1}{4} [e^{u}]_{- 8}^{- 4})) : \frac{3}{4} [e^{u}]_{- 8}^{- 4}

= \frac{3}{4} [e^{u}]_{- 8}^{- 4}

Compute the boundaries:

\frac{1}{e ^{4}} - \frac{1}{e ^{8}}

= \frac{3}{4} (\frac{1}{e ^{4}} - \frac{1}{e ^{8}})

Simplify

= \frac{3 ( e ^{4} - 1 )}{4 e ^{8}}

y^{^{'}} = - 0.5 y + x

\int x^{4} sin (x) d x

\int \frac{y}{2} d y

f (x) = 5 x 2

What is the integral from 1 to 2 of 3/(e^{4t)} ?
The integral from 1 to 2 of 3/(e^{4t)} is (3(e^4-1))/(4e^8)