What is the integral from-1 to 1 of x/(sqrt(5-4x)) ?

The integral from-1 to 1 of x/(sqrt(5-4x)) is 1/6

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integral from-1 to 1 of x/(sqrt(5-4x))

\int_{- 1}^{1} \frac{x}{5 - 4 x} d x

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

\int_{- 1}^{1} \frac{x}{5 - 4 x} d x

Apply u-substitution

: \int_{3}^{1} \frac{u ^{2} - 5}{8} d u

= \int_{3}^{1} \frac{u ^{2} - 5}{8} d u

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

= - \int_{1}^{3} \frac{u ^{2} - 5}{8} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int_{1}^{3} u^{2} - 5 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{1}{8} (\int_{1}^{3} u^{2} d u - \int_{1}^{3} 5 d u)

\int_{1}^{3} u^{2} d u = \frac{26}{3}

\int_{1}^{3} 5 d u = 10

= - \frac{1}{8} (\frac{26}{3} - 10)

Simplify

- \frac{1}{8} (\frac{26}{3} - 10) : \frac{1}{6}

= \frac{1}{6}

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

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 1

f (x) = \frac{1}{x ^{\frac{2}{3}}}

\int \frac{4 x ^{3} + 3 x + 3}{x} d x

x \to 7 lim (sin (\frac{π}{x}))

What is the integral from-1 to 1 of x/(sqrt(5-4x)) ?
The integral from-1 to 1 of x/(sqrt(5-4x)) is 1/6