What is the integral from-2 to 0 of-7e^{-6t-3} ?

The integral from-2 to 0 of-7e^{-6t-3} is -(7(e^{12}-1))/(6e^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from-2 to 0 of-7e^{-6t-3}

\int_{- 2}^{0} - 7 e^{- 6 t - 3} d t

- \frac{7 ( e ^{12} - 1 )}{6 e ^{3}}

Decimal

- 9453.53983 \dots

Solution steps

\int_{- 2}^{0} - 7 e^{- 6 t - 3} d t

Take the constant out:

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

= - 7 \cdot \int_{- 2}^{0} e^{- 6 t - 3} d t

Apply u-substitution

= - 7 \cdot \int_{9}^{- 3} - \frac{1}{6} e^{u} d u

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

= - 7 (- \int_{- 3}^{9} - \frac{1}{6} 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}{6} \cdot \int_{- 3}^{9} e^{u} d u))

Use the common integral:

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

= - 7 (- (- \frac{1}{6} [e^{u}]_{- 3}^{9}))

Simplify

- 7 (- (- \frac{1}{6} [e^{u}]_{- 3}^{9})) : - \frac{7}{6} [e^{u}]_{- 3}^{9}

= - \frac{7}{6} [e^{u}]_{- 3}^{9}

Compute the boundaries:

e^{9} - \frac{1}{e ^{3}}

= - \frac{7}{6} (e^{9} - \frac{1}{e ^{3}})

Simplify

= - \frac{7 ( e ^{12} - 1 )}{6 e ^{3}}

\int_{1}^{2} \frac{1}{x + e} d x

\int_{- 3}^{4} 1 d x

\int_{0}^{2 π} 4 sin^{2} (x) d x

\int_{- π}^{0} - x d x

\int_{0}^{1} x (x^{2} + 2)^{6} d x

What is the integral from-2 to 0 of-7e^{-6t-3} ?
The integral from-2 to 0 of-7e^{-6t-3} is -(7(e^{12}-1))/(6e^3)