What is the integral from-32 to 0 of 1/(sqrt(4-x)) ?

The integral from-32 to 0 of 1/(sqrt(4-x)) is 8

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integral from-32 to 0 of 1/(sqrt(4-x))

\int_{- 32}^{0} \frac{1}{4 - x} d x

8

Solution steps

\int_{- 32}^{0} \frac{1}{4 - x} d x

Apply u-substitution

: \int_{36}^{4} - \frac{1}{u} d u

= \int_{36}^{4} - \frac{1}{u} d u

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

= - \int_{4}^{36} - \frac{1}{u} d u

Take the constant out:

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

= - (- \int_{4}^{36} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{4}^{36}

= - (- [2 u^{\frac{1}{2}}]_{4}^{36})

Simplify

- (- [2 u^{\frac{1}{2}}]_{4}^{36}) : [2 u]_{4}^{36}

= [2 u]_{4}^{36}

Compute the boundaries:

8

= 8

\frac{d}{d x} (\frac{3 cos ( x )}{3 x ^{3} + 2 x ^{2} - x + 4})

\int 4 cos^{2} (3 x) d x

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

sin (2 x)

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

What is the integral from-32 to 0 of 1/(sqrt(4-x)) ?
The integral from-32 to 0 of 1/(sqrt(4-x)) is 8