What is the integral from 0 to 0.6 of (x^2)/(sqrt(9-25x^2)) ?

The integral from 0 to 0.6 of (x^2)/(sqrt(9-25x^2)) is (9pi)/(500)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 0.6 of (x^2)/(sqrt(9-25x^2))

\int_{0}^{0.6} \frac{x ^{2}}{9 - 25 x ^{2}} d x

\frac{9 π}{500}

Decimal

0.05654 \dots

Solution steps

\int_{0}^{0.6} \frac{x ^{2}}{9 - 25 x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} \frac{9 sin ^{2} ( u )}{125} d u

= \int_{0}^{\frac{π}{2}} \frac{9 sin ^{2} ( u )}{125} d u

Take the constant out:

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

= \frac{9}{125} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) d u

Rewrite using trig identities

= \frac{9}{125} \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 2 u )}{2} d u

Take the constant out:

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

= \frac{9}{125} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{9}{125} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= \frac{9}{125} \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

\frac{9}{125} \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{9 π}{500}

= \frac{9 π}{500}

2 x^{4} - 9 x^{2}

x y^{^{'}} + y = 3

t y^{^{'}} + (t + 1) y = t, y (ln (3)) = 1

x \to 0 lim (x^{4} (cos (\frac{2}{x})))

\int \frac{4}{5} d x

What is the integral from 0 to 0.6 of (x^2)/(sqrt(9-25x^2)) ?
The integral from 0 to 0.6 of (x^2)/(sqrt(9-25x^2)) is (9pi)/(500)