What is the integral from-21 to 0 of 1/(sqrt(4-x)) ?

The integral from-21 to 0 of 1/(sqrt(4-x)) is 6

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integral from-21 to 0 of 1/(sqrt(4-x))

\int_{- 21}^{0} \frac{1}{4 - x} d x

6

Solution steps

\int_{- 21}^{0} \frac{1}{4 - x} d x

Apply u-substitution

: \int_{25}^{4} - \frac{1}{u} d u

= \int_{25}^{4} - \frac{1}{u} d u

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

= - \int_{4}^{25} - \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}^{25} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{4}^{25}

= - (- [2 u^{\frac{1}{2}}]_{4}^{25})

Simplify

- (- [2 u^{\frac{1}{2}}]_{4}^{25}) : [2 u]_{4}^{25}

= [2 u]_{4}^{25}

Compute the boundaries:

6

= 6

\int 8 cos (x) d x

\int_{0}^{2} 10 x^{3} x^{2} + 4 d x

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = 2 x - 17

\frac{\partial}{\partial y} (x tan (y))

y = - 2 x^{2} - x + 6

What is the integral from-21 to 0 of 1/(sqrt(4-x)) ?
The integral from-21 to 0 of 1/(sqrt(4-x)) is 6