What is the integral from-1 to 1 of (2sqrt(5-4x))/3 ?

The integral from-1 to 1 of (2sqrt(5-4x))/3 is 26/9

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integral from-1 to 1 of (2sqrt(5-4x))/3

\int_{- 1}^{1} \frac{2 5 - 4 x}{3} d x

\frac{26}{9}

Decimal

2.88888 \dots

Solution steps

\int_{- 1}^{1} \frac{2 5 - 4 x}{3} d x

Take the constant out:

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

= \frac{2}{3} \cdot \int_{- 1}^{1} 5 - 4 x d x

Apply u-substitution

= \frac{2}{3} \cdot \int_{9}^{1} - \frac{u}{4} d u

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

= \frac{2}{3} (- \int_{1}^{9} - \frac{u}{4} d u)

Take the constant out:

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

= \frac{2}{3} (- (- \frac{1}{4} \cdot \int_{1}^{9} u d u))

Apply the Power Rule

= \frac{2}{3} (- (- \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}))

Simplify

\frac{2}{3} (- (- \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9})) : \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

\frac{52}{3}

= \frac{1}{6} \cdot \frac{52}{3}

Simplify

= \frac{26}{9}

in v erse l a pl a ce \frac{x + 1}{x ^{2} + x + 10}

\frac{d}{d x} (f (x) (x^{3} + x - 3))

y^{^{'}} - e^{3 x + y} = 0

n = 0 \sum \infty \frac{2}{n ^{2} - 1}

\int_{0}^{\infty} f (a x) d x

What is the integral from-1 to 1 of (2sqrt(5-4x))/3 ?
The integral from-1 to 1 of (2sqrt(5-4x))/3 is 26/9