What is the integral from-12 to 0 of 1/(sqrt(4-x)) ?

The integral from-12 to 0 of 1/(sqrt(4-x)) is 4

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integral from-12 to 0 of 1/(sqrt(4-x))

\int_{- 12}^{0} \frac{1}{4 - x} d x

4

Solution steps

\int_{- 12}^{0} \frac{1}{4 - x} d x

Apply u-substitution

: \int_{16}^{4} - \frac{1}{u} d u

= \int_{16}^{4} - \frac{1}{u} d u

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

= - \int_{4}^{16} - \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}^{16} \frac{1}{u} d u)

Apply the Power Rule

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

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

Simplify

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

= [2 u]_{4}^{16}

Compute the boundaries:

4

= 4

\frac{\partial}{\partial x} (x \cdot arctan (\frac{y}{x}))

\int \frac{x ^{2}}{( 15 + 6 x - 9 x ^{2} )} d x

n = 1 \sum \infty \frac{1}{3 ^{\frac{1}{n}}}

x \to 3 lim (6 x - 7)

f (x) = x^{2} - 8, at x = 7

What is the integral from-12 to 0 of 1/(sqrt(4-x)) ?
The integral from-12 to 0 of 1/(sqrt(4-x)) is 4