What is the integral from 1 to 24 of sqrt(1+x) ?

The integral from 1 to 24 of sqrt(1+x) is (250-4sqrt(2))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 24 of sqrt(1+x)

\int_{1}^{24} 1 + x d x

\frac{250 - 4 2}{3}

Decimal Notation

81.44771 \dots

Solution steps

\int_{1}^{24} 1 + x d x

Apply u-substitution

: \int_{2}^{25} u d u

= \int_{2}^{25} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{25}

= [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{25}

Compute the boundaries:

\frac{250}{3} - \frac{4 2}{3}

= \frac{250}{3} - \frac{4 2}{3}

Simplify

= \frac{250 - 4 2}{3}

\int_{z}^{y} \frac{1}{x} d x

\int_{y^{2}}^{y} y d y

\int_{0}^{0.5} 96 x^{2} y^{3} d x

\int_{\frac{2}{3}}^{10} (11 - x)^{\frac{1}{2}} d x

\int_{0}^{2 π} - sin^{3} (x) d x

What is the integral from 1 to 24 of sqrt(1+x) ?
The integral from 1 to 24 of sqrt(1+x) is (250-4sqrt(2))/3