What is the integral from 1 to 16 of e^{sqrt(x)} ?

The integral from 1 to 16 of e^{sqrt(x)} is 6e^4

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integral from 1 to 16 of e^{sqrt(x)}

\int_{1}^{16} e^{x} d x

6 e^{4}

Decimal

327.58890 \dots

Solution steps

\int_{1}^{16} e^{x} d x

Apply u-substitution

= \int_{1}^{4} 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}^{4} e^{u} u d u

Apply Integration By Parts

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

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

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

Compute the boundaries:

3 e^{4}

= 2 \cdot 3 e^{4}

Simplify

= 6 e^{4}

y^{^{''}} + 4 y^{^{'}} + 13 y = 3 cos (2 x)

\int \frac{1}{y ^{5}} d y

y^{^{''}} - y = 0, y (0) = 0, y (1) = 1

\int \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

x \to - 1 lim (- 12 x^{3})

What is the integral from 1 to 16 of e^{sqrt(x)} ?
The integral from 1 to 16 of e^{sqrt(x)} is 6e^4