What is the integral from-1 to 3 of 3/(sqrt(6-2x)) ?

The integral from-1 to 3 of 3/(sqrt(6-2x)) is 6sqrt(2)

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integral from-1 to 3 of 3/(sqrt(6-2x))

\int_{- 1}^{3} \frac{3}{6 - 2 x} d x

62

Decimal

8.48528 \dots

Solution steps

\int_{- 1}^{3} \frac{3}{6 - 2 x} d x

Take the constant out:

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

= 3 \cdot \int_{- 1}^{3} \frac{1}{6 - 2 x} d x

Apply u-substitution

: \int_{8}^{0} - \frac{1}{2 u} d u

= 3 \cdot \int_{8}^{0} - \frac{1}{2 u} d u

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

= 3 (- \int_{0}^{8} - \frac{1}{2 u} d u)

Take the constant out:

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

= 3 (- (- \frac{1}{2} \cdot \int_{0}^{8} \frac{1}{u} d u))

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{8}

= 3 (- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{8}))

Simplify

3 (- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{8})) : \frac{3}{2} [2 u]_{0}^{8}

= \frac{3}{2} [2 u]_{0}^{8}

Compute the boundaries:

42

= \frac{3}{2} \cdot 42

Simplify

= 62

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

\int_{1}^{9} x^{3} d x

\int_{0.3}^{0.6} x d x

\int_{1}^{9} \frac{1}{5 x + 4} d x

\int_{0}^{4} 2 x - \frac{1}{4} x^{2} d x

What is the integral from-1 to 3 of 3/(sqrt(6-2x)) ?
The integral from-1 to 3 of 3/(sqrt(6-2x)) is 6sqrt(2)