What is the integral from 1 to 4 of-3e^{sqrt(x)} ?

The integral from 1 to 4 of-3e^{sqrt(x)} is -6e^2

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integral from 1 to 4 of-3e^{sqrt(x)}

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

- 6 e^{2}

Decimal Notation

- 44.33433 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - 3 \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

= - 3 \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}

= - 3 \cdot 2 [e^{u} u - \int e^{u} d u]_{1}^{2}

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

= - 3 \cdot 2 [e^{u} u - e^{u}]_{1}^{2}

Simplify

= - 6 [e^{u} u - e^{u}]_{1}^{2}

Compute the boundaries:

e^{2}

= - 6 e^{2}

\int 20 sec (4 x) tan (4 x) d x

\int \frac{3 x ^{2}}{25 - x ^{2}} d x

x \to 7 lim (cos (\frac{π x}{3}))

y = 3 x

x \to 3 lim (2 x^{3} + 4 x - 2)

What is the integral from 1 to 4 of-3e^{sqrt(x)} ?
The integral from 1 to 4 of-3e^{sqrt(x)} is -6e^2