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}

\frac{d}{d x} (tan (π))

x = 1, y = 6 x (2 - x^{2})^{2}

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

x \to a + lim (x^{n})

\int \frac{x ^{6} + 4 x ^{3} + 4}{x ^{3} - 4 x ^{2}} d x

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