What is the integral from 1 to 0 of (-9sqrt(x)) ?

The integral from 1 to 0 of (-9sqrt(x)) is 6

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integral from 1 to 0 of (-9sqrt(x))

\int_{1}^{0} (- 9 x) d x

6

Solution steps

\int_{1}^{0} - 9 x d x

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

= - \int_{0}^{1} - 9 x d x

Take the constant out:

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

= - (- 9 \cdot \int_{0}^{1} x d x)

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{1}

= - (- 9 [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{1})

Simplify

= 9 [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{1}

Compute the boundaries:

\frac{2}{3}

= 9 \cdot \frac{2}{3}

Simplify

= 6

\frac{\partial}{\partial t} (ln (x + 2 c t))

\int_{- 1}^{2} ∣ 2 x - 3 ∣ d x

y^{^{''}} + 36 y = 0

\int (cos (x) sin^{2} (x)) d x

\frac{d}{d x} (\frac{2 x ^{2} - 1}{x ^{2} - 3})

What is the integral from 1 to 0 of (-9sqrt(x)) ?
The integral from 1 to 0 of (-9sqrt(x)) is 6