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

y^{^{''}} - 4 y^{^{'}} - 12 y = t e^{4 t}

e^{x} \frac{( 2 x + 5 ) ^{2}}{x - 4}

\int 3 sin^{3} (x) cos (x) d x

x \to 3 lim (5 (- 2 x + 3))

\frac{sin ( x )}{cos ( ln ( - x ) )}

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