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

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

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

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

2 e^{2}

Decimal

14.77811 \dots

Solution steps

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

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

e^{2}

= 2 e^{2}

\int x^{2} - 16 d x

f (x) = x^{3} - 2 x + 1, (3, 22)

f (x) = \frac{1 + ln ( x )}{x}, at x = e

y = \frac{x}{x + 7}

64 x - 0.02 x^{2}

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