What is the integral from-8 to 0 of (4-9x)^{-0.5} ?

The integral from-8 to 0 of (4-9x)^{-0.5} is (4sqrt(19)-4)/9

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integral from-8 to 0 of (4-9x)^{-0.5}

\int_{- 8}^{0} (4 - 9 x)^{- 0.5} d x

\frac{4 19 - 4}{9}

Decimal Notation

1.49284 \dots

Solution steps

\int_{- 8}^{0} (4 - 9 x)^{- 0.5} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int_{- 8}^{0} \frac{1}{4 - 9 x} d x

Apply u-substitution

: \int_{76}^{4} - \frac{1}{9 u} d u

= \int_{76}^{4} - \frac{1}{9 u} d u

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

= - \int_{4}^{76} - \frac{1}{9 u} d u

Take the constant out:

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

= - (- \frac{1}{9} \cdot \int_{4}^{76} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{4}^{76}

= - (- \frac{1}{9} [2 u^{\frac{1}{2}}]_{4}^{76})

Simplify

- (- \frac{1}{9} [2 u^{\frac{1}{2}}]_{4}^{76}) : \frac{1}{9} [2 u]_{4}^{76}

= \frac{1}{9} [2 u]_{4}^{76}

Compute the boundaries:

419 - 4

= \frac{1}{9} (419 - 4)

Simplify

= \frac{4 19 - 4}{9}

(x - 1)^{3}, (x - 1)

\frac{\partial}{\partial x} (\frac{1}{2} ln (x^{2} d))

\frac{d}{d x} (csc (x))

\int 15 cos (15 x) d x

f (x) = 9 x - 10 x^{\frac{1}{4}}

What is the integral from-8 to 0 of (4-9x)^{-0.5} ?
The integral from-8 to 0 of (4-9x)^{-0.5} is (4sqrt(19)-4)/9