What is the integral from 1 to 4 of-9e^{sqrt(x)} ?

The integral from 1 to 4 of-9e^{sqrt(x)} is -18e^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 4 of-9e^{sqrt(x)}

\int_{1}^{4} - 9 e^{x} d x

- 18 e^{2}

Decimal Notation

- 133.00300 \dots

Solution steps

\int_{1}^{4} - 9 e^{x} d x

Take the constant out:

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

= - 9 \cdot \int_{1}^{4} e^{x} d x

Apply u-substitution

: \int_{1}^{2} e^{u} \cdot 2 u d u

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

Take the constant out:

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

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

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{1}^{2}

= - 9 \cdot 2 [e^{u} u - \int e^{u} d u]_{1}^{2}

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

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

Simplify

= - 18 [e^{u} u - e^{u}]_{1}^{2}

Compute the boundaries:

e^{2}

= - 18 e^{2}

(- 3, 7), (2, - 6)

\int 4 x e^{6 x} d x

- \frac{1}{2} x^{2} + \frac{5}{2} x - 2, [- 2, 7]

f (x) = \frac{1 + x}{( 2 x ^{2} + 1 ) ^{2}}

y^{^{'}} + 2 y = 0, y (1) = 1

What is the integral from 1 to 4 of-9e^{sqrt(x)} ?
The integral from 1 to 4 of-9e^{sqrt(x)} is -18e^2