What is the integral from-2 to 2 of 1/(sqrt(4-x)) ?

The integral from-2 to 2 of 1/(sqrt(4-x)) is 2sqrt(6)-2sqrt(2)

Popular >

integral from-2 to 2 of 1/(sqrt(4-x))

\int_{- 2}^{2} \frac{1}{4 - x} d x

26 - 22

Decimal

2.07055 \dots

Solution steps

\int_{- 2}^{2} \frac{1}{4 - x} d x

Apply u-substitution

: \int_{6}^{2} - \frac{1}{u} d u

= \int_{6}^{2} - \frac{1}{u} d u

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

= - \int_{2}^{6} - \frac{1}{u} d u

Take the constant out:

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

= - (- \int_{2}^{6} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{2}^{6}

= - (- [2 u^{\frac{1}{2}}]_{2}^{6})

Simplify

- (- [2 u^{\frac{1}{2}}]_{2}^{6}) : [2 u]_{2}^{6}

= [2 u]_{2}^{6}

Compute the boundaries:

26 - 22

= 26 - 22

\int_{0}^{1} sin^{2} (3 x) d x

\int_{0}^{1} 2 x + 5 d x

\int_{0}^{π} \frac{1}{2} cos (x) d x

\int_{0}^{x} \frac{1}{( 1 - x ) ^{1.5}} d x

\int_{0}^{2 π} 18 cos^{2} (t) d t

What is the integral from-2 to 2 of 1/(sqrt(4-x)) ?
The integral from-2 to 2 of 1/(sqrt(4-x)) is 2sqrt(6)-2sqrt(2)