What is the integral from 0 to 0.4 of ((x^2))/(sqrt(4-25x^2)) ?

The integral from 0 to 0.4 of ((x^2))/(sqrt(4-25x^2)) is pi/(125)

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integral from 0 to 0.4 of ((x^2))/(sqrt(4-25x^2))

\int_{0}^{0.4} \frac{( x ^{2} )}{4 - 25 x ^{2}} d x

\frac{π}{125}

Decimal Notation

0.02513 \dots

Solution steps

\int_{0}^{0.4} \frac{x ^{2}}{4 - 25 x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} \frac{4 sin ^{2} ( u )}{125} d u

= \int_{0}^{\frac{π}{2}} \frac{4 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{4}{125} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) d u

Rewrite using trig identities

= \frac{4}{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{4}{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{4}{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{4}{125} \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

\frac{4}{125} \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{π}{125}

= \frac{π}{125}

\int (sin (x)) (cos (x)) d x

\frac{d}{d x} ((x^{5} - 9) x)

x = - 3, x = 1, y = x^{3} + 1, y = 0

\int_{2}^{\infty} x e^{- 2 x} d x

\frac{\partial}{\partial x} (y x - y - x)

What is the integral from 0 to 0.4 of ((x^2))/(sqrt(4-25x^2)) ?
The integral from 0 to 0.4 of ((x^2))/(sqrt(4-25x^2)) is pi/(125)