What is the integral from 0 to 4 of xsqrt(17-x^2) ?

The integral from 0 to 4 of xsqrt(17-x^2) is (17sqrt(17)-1)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of xsqrt(17-x^2)

\int_{0}^{4} x 17 - x^{2} d x

\frac{17 17 - 1}{3}

Decimal Notation

23.03093 \dots

Solution steps

\int_{0}^{4} x 17 - x^{2} d x

Apply u-substitution

: \int_{17}^{1} - \frac{u}{2} d u

= \int_{17}^{1} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{17} - \frac{u}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{2} \cdot \int_{1}^{17} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Compute the boundaries:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{2} (\frac{34 17}{3} - \frac{2}{3})

Simplify

= \frac{17 17 - 1}{3}

\int_{0}^{1} \frac{x ^{4} + 6}{x} d x

tan^{4} (2 x - 1)^{3}

\int \frac{x}{x + 1 - x + 1} d x

x \to 1 + lim (\frac{3}{ln ( x )} - \frac{2}{x - 1})

y = \frac{4}{x}, (5, \frac{4}{5})

What is the integral from 0 to 4 of xsqrt(17-x^2) ?
The integral from 0 to 4 of xsqrt(17-x^2) is (17sqrt(17)-1)/3