What is the integral from 0 to 1 of e^{x^{1/2}} ?

The integral from 0 to 1 of e^{x^{1/2}} is 2

Popular >

integral from 0 to 1 of e^{x^{1/2}}

\int_{0}^{1} e^{x^{\frac{1}{2}}} d x

2

Solution steps

\int_{0}^{1} e^{x^{\frac{1}{2}}} d x

Apply u-substitution

: \int_{0}^{1} e^{u} \cdot 2 u d u

= \int_{0}^{1} 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_{0}^{1} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{1}

= 2 [e^{u} u - \int e^{u} d u]_{0}^{1}

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

= 2 [e^{u} u - e^{u}]_{0}^{1}

Compute the boundaries:

1

= 2 \cdot 1

Simplify

= 2

\int_{- 2}^{4} - x + 4 d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} cos^{2} (x) d x

\int_{- 1}^{3} (2 x + 2) - (x^{2} - 1) d x

\int_{1}^{7} 4 d x

\int_{- 1}^{4} 16 - x^{2} d x

What is the integral from 0 to 1 of e^{x^{1/2}} ?
The integral from 0 to 1 of e^{x^{1/2}} is 2