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

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

integral from 1 to 4 of-4e^{sqrt(x)}

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

- 8 e^{2}

Decimal

- 59.11244 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= - 4 \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

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

Apply Integration By Parts

= - 4 \cdot 2 [e^{u} u - \int e^{u} d u]_{1}^{2}

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

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

Simplify

= - 8 [e^{u} u - e^{u}]_{1}^{2}

Compute the boundaries:

e^{2}

= - 8 e^{2}

d er i v a t i v e f (x) = (x - 2) (x - 1) (x + 1) (x + 2)

\frac{d}{d x} (x^{4} + \frac{1}{x ^{4}})

d er i v a t i v e \frac{s - 4}{s + 2}

\frac{d}{d x} (arccos (\frac{1}{2} x))

im pl i c i t e^{x y} + y^{3} = 2 arcsin (x) + ln (y)

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