What is the integral from-6 to 0 of sqrt(4-2x) ?

The integral from-6 to 0 of sqrt(4-2x) is 56/3

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integral from-6 to 0 of sqrt(4-2x)

\int_{- 6}^{0} 4 - 2 x d x

\frac{56}{3}

Decimal

18.66666 \dots

Solution steps

\int_{- 6}^{0} 4 - 2 x d x

Apply u-substitution

: \int_{16}^{4} - \frac{u}{2} d u

= \int_{16}^{4} - \frac{u}{2} d u

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

= - \int_{4}^{16} - \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_{4}^{16} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{16}

Compute the boundaries:

\frac{112}{3}

= \frac{1}{2} \cdot \frac{112}{3}

Simplify

= \frac{56}{3}

\int_{0}^{1} 2 π x e^{x} d x

\int_{0}^{2} \frac{1}{4 - x} d x

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

\int_{1}^{2} \frac{1}{( x - 1 ) ( x ^{2} + 1 )} d x

\int_{0}^{1} (\frac{6}{x ^{2} + 1}) d x

What is the integral from-6 to 0 of sqrt(4-2x) ?
The integral from-6 to 0 of sqrt(4-2x) is 56/3